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GRADE 10 PRACTICE

Physics Problems: Molecular kinetic theory, Grade 10

Gas particles, temperature and gas laws.

24 problems across 4 sub-topics
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Problem statements

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24 problems
Position in this set: 1 from 24

Problem #71

Molecular kinetic theory: particles and their movement

Determine molar mass N_2O_5, mass of one molecule, number of molecules in 1\,\text{кг} substances and mass 2\cdot10^{20} molecules.

Answer options

  1. \left(1080\,\text{г/моль};\ 1{,}8 \cdot 10^{-24}\,\text{кг};\ 5{,}6 \cdot 10^{25}\,\text{молекул};\ 3{,}6 \cdot 10^{-4}\,\text{кг}\right)
  2. \left(108\,\text{г/моль};\ 1{,}8 \cdot 10^{-25}\,\text{кг};\ 5{,}6 \cdot 10^{24}\,\text{молекул};\ 3{,}6 \cdot 10^{-5}\,\text{кг}\right)
  3. \left(54\,\text{г/моль};\ 9 \cdot 10^{-26}\,\text{кг};\ 2{,}8 \cdot 10^{24}\,\text{молекул};\ 1{,}8 \cdot 10^{-5}\,\text{кг}\right)
  4. \left(216\,\text{г/моль};\ 3{,}6 \cdot 10^{-25}\,\text{кг};\ 1{,}12 \cdot 10^{25}\,\text{молекул};\ 7{,}2 \cdot 10^{-5}\,\text{кг}\right)
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Given
\text{Substance}
\mathrm{N_2O_5}
m
1\,\text{кг}
N_2
2\cdot 10^{20}
N_A
6\cdot 10^{23}\,\tfrac{1}{\text{моль}}
Find

M_rm_0Nm_2

  1. M_r = 2\cdot 14 + 5\cdot 16M_r = 108\ \Rightarrow\ M = 108\,\tfrac{\text{г}}{\text{моль}}
  2. m_0 = \dfrac{M}{N_A}m_0 = \dfrac{108\,\tfrac{\text{г}}{\text{моль}}}{6\cdot 10^{23}\,\tfrac{1}{\text{моль}}}m_0 = 1{,}8\cdot 10^{-22}\,\text{г} = 1{,}8\cdot 10^{-25}\,\text{кг}
  3. N = \dfrac{m}{M}N_AN = \dfrac{1000\,\text{г}\cdot 6\cdot 10^{23}}{108\,\tfrac{\text{г}}{\text{моль}}}N = 5{,}6\cdot 10^{24}
  4. m_2 = N_2 m_0m_2 = 2\cdot 10^{20}\cdot 1{,}8\cdot 10^{-25}m_2 = 3{,}6\cdot 10^{-5}\,\text{кг}

AnswerM_r = 108m_0 = 1{,}8\cdot 10^{-25}\,\text{кг}N = 5{,}6\cdot 10^{24}m_2 = 3{,}6\cdot 10^{-5}\,\text{кг}

Position in this set: 2 from 24

Problem #72

Molecular kinetic theory: particles and their movement

Available 93\,\text{г} carbonic acid H_2CO_3. Determine the relative molecular mass, molar mass, amount of substance, number of molecules, mass of one molecule, number of atoms of each element, and the total number of protons, neutrons, and electrons. Use mass numbers 1, 12, and 16 for hydrogen, carbon, and oxygen.

Answer options

  1. \left(124\,\text{безразм.};\ 0{,}124\,\text{кг/моль};\ 3\,\text{моль};\ 1{,}8 \cdot 10^{24}\,\text{молекул};\ 2{,}06 \cdot 10^{-25}\,\text{кг};\ 3{,}6 \cdot 10^{24}\,\text{атомов H};\ 1{,}8 \cdot 10^{24}\,\text{атомов C};\ 5{,}4 \cdot 10^{24}\,\text{атомов O};\ 1{,}692 \cdot 10^{26}\,\text{частиц}\right)
  2. \left(620\,\text{безразм.};\ 0{,}62\,\text{кг/моль};\ 15\,\text{моль};\ 9 \cdot 10^{24}\,\text{молекул};\ 1{,}03 \cdot 10^{-24}\,\text{кг};\ 1{,}8 \cdot 10^{25}\,\text{атомов H};\ 9 \cdot 10^{24}\,\text{атомов C};\ 2{,}7 \cdot 10^{25}\,\text{атомов O};\ 8{,}46 \cdot 10^{26}\,\text{частиц}\right)
  3. \left(62\,\text{безразм.};\ 0{,}062\,\text{кг/моль};\ 1{,}5\,\text{моль};\ 9 \cdot 10^{23}\,\text{молекул};\ 1{,}03 \cdot 10^{-25}\,\text{кг};\ 1{,}8 \cdot 10^{24}\,\text{атомов H};\ 9 \cdot 10^{23}\,\text{атомов C};\ 2{,}7 \cdot 10^{24}\,\text{атомов O};\ 8{,}46 \cdot 10^{25}\,\text{частиц}\right)
  4. \left(31\,\text{безразм.};\ 0{,}031\,\text{кг/моль};\ 0{,}75\,\text{моль};\ 4{,}5 \cdot 10^{23}\,\text{молекул};\ 5{,}15 \cdot 10^{-26}\,\text{кг};\ 9 \cdot 10^{23}\,\text{атомов H};\ 4{,}5 \cdot 10^{23}\,\text{атомов C};\ 1{,}35 \cdot 10^{24}\,\text{атомов O};\ 4{,}23 \cdot 10^{25}\,\text{частиц}\right)
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Given
\text{Substance}
\mathrm{H_2CO_3}
m
93\,\text{г}
N_A
6\cdot 10^{23}\,\tfrac{1}{\text{моль}}
Find

M_rM\nuNm_0N_{\mathrm{H}}N_{\mathrm{C}}N_{\mathrm{O}}N_{\text{частиц}}

  1. M_r = 2\cdot 1 + 12 + 3\cdot 16M = 62\ \Rightarrow\ M = 62\,\tfrac{\text{г}}{\text{моль}} = 0{,}062\,\tfrac{\text{кг}}{\text{моль}}
  2. \nu = \dfrac{m}{M}\nu = \dfrac{93}{62}\nu = 1{,}5\,\text{моль}
  3. N = \nu N_AN = 1{,}5\cdot 6\cdot 10^{23}N = 9\cdot 10^{23}
  4. m_0 = \dfrac{M}{N_A}m_0 = \dfrac{0{,}062}{6\cdot10^{23}}m_0 = 1{,}03\cdot10^{-25}\,\text{кг}
  5. N_{\mathrm{H}} = 2N,\quad N_{\mathrm{C}} = N,\quad N_{\mathrm{O}} = 3NN_{\mathrm{H}} = 1{,}8\cdot 10^{24};\quad N_{\mathrm{C}} = 9\cdot 10^{23};\quad N_{\mathrm{O}} = 2{,}7\cdot 10^{24}

    One molecule contains two hydrogen atoms, one carbon and three oxygen atoms.

  6. N_{\text{частиц}} = N\left(2\cdot 2 + 18 + 3\cdot 24\right)N_{\text{частиц}} = 9\cdot 10^{23}\cdot 94N_{\text{частиц}} = 8{,}46\cdot 10^{25}

    There are 2 particles in a hydrogen atom, 18 in carbon-12, and 24 in oxygen-16.

AnswerM_r = 62M = 0{,}062\,\tfrac{\text{кг}}{\text{моль}}\nu = 1{,}5\,\text{моль}N = 9\cdot 10^{23}m_0 = 1{,}03\cdot 10^{-25}\,\text{кг}N_{\mathrm{H}} = 1{,}8\cdot 10^{24}N_{\mathrm{C}} = 9\cdot 10^{23}N_{\mathrm{O}} = 2{,}7\cdot 10^{24}N_{\text{частиц}} = 8{,}46\cdot 10^{25}

Position in this set: 3 from 24

Problem #73

Molecular kinetic theory: particles and their movement

Into the lake area 10\,\text{км}^{2} and average depth 2\,\text{м} contributed 1\,\text{мг} sodium chloride NaCl. How many formula units of NaCl are there per glass of water? 300\,\text{мл} after complete dissolution and uniform distribution? Accept M=58{,}5\,\text{г/моль}, N_A=6{,}02\cdot10^{23}\,\text{моль}^{-1}. The formula unit here refers to one pair of Na⁺ and Cl⁻ ions.

Answer options

  1. 7{,}7 \cdot 10^{7}\,\text{формульных единиц}
  2. 3{,}08 \cdot 10^{8}\,\text{формульных единиц}
  3. 1{,}54 \cdot 10^{9}\,\text{формульных единиц}
  4. 1{,}54 \cdot 10^{8}\,\text{формульных единиц}
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Given
S
10\,\text{км}^{2}10^7\,\text{м}^{2}
h
2\,\text{м}
m
1\,\text{мг}10^{-6}\,\text{кг}
V_c
300\,\text{мл}3\cdot10^{-4}\,\text{м}^{3}
M
58{,}5\,\text{г/моль}0{,}0585\,\text{кг/моль}
N_A
6{,}02\cdot10^{23}\,\text{моль}^{-1}
Find

N_c

  1. V=ShV = 10^7\cdot2V = 2\cdot10^7\,\text{м}^{3}
  2. N_c=\dfrac{mN_A}{M}\dfrac{V_c}{V}N_c = \dfrac{10^{-6}\cdot6{,}02\cdot10^{23}}{0{,}0585}\cdot\dfrac{3\cdot10^{-4}}{2\cdot10^7}N_c = 1{,}54\cdot10^8

    The proportion of dissolved salt in a glass is equal to the proportion of the volume of the glass in the volume of the lake. NaCl dissociates in water; We count pairs of ions, not molecules.

AnswerN_c=1{,}54\cdot10^8\,\text{формульных единиц}

Position in this set: 4 from 24

Problem #74

Molecular kinetic theory: particles and their movement

A glass of water evaporated in 30 days. How many molecules were released from the water every second? Take a lot of water 200\,\text{г}.

Answer options

  1. 2{,}6 \cdot 10^{18}\,\text{молекул/с}
  2. 1{,}3 \cdot 10^{18}\,\text{молекул/с}
  3. 5{,}2 \cdot 10^{18}\,\text{молекул/с}
  4. 2{,}6 \cdot 10^{19}\,\text{молекул/с}
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Given
t
30\,\text{сут}2\,592\,000\,\text{с}
m
200\,\text{г}
M
18\,\tfrac{\text{г}}{\text{моль}}
N_A
6\cdot 10^{23}\,\tfrac{1}{\text{моль}}
Find

N'

  1. N = \dfrac{m}{M}N_AN = \dfrac{200\cdot 6\cdot 10^{23}}{18}N = 6{,}7\cdot 10^{24}
  2. N' = \dfrac{N}{t}N' = \dfrac{6{,}7\cdot 10^{24}}{2\,592\,000}N' = 2{,}6\cdot 10^{18}\,\tfrac{1}{\text{с}}

AnswerN' = 2{,}6\cdot 10^{18}\,\tfrac{1}{\text{с}}

Position in this set: 5 from 24

Problem #75

Molecular kinetic theory: particles and their movement

What pressure does the gas exert? NO_2, if the concentration of molecules 3\cdot10^{25}\,\text{м}^{-3}, and the mean square speed 400\,\text{м/с}?

Answer options

  1. 1{,}2 \cdot 10^{6}\,\text{Па}
  2. 1{,}2 \cdot 10^{5}\,\text{Па}
  3. 60000\,\text{Па}
  4. 2{,}4 \cdot 10^{5}\,\text{Па}
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Given
\text{газ}
\mathrm{NO_2}
M
46\,\tfrac{\text{г}}{\text{моль}}
n
3\cdot 10^{25}\,\tfrac{1}{\text{м}^{3}}
\bar v
400\,\text{м/с}
Find

p

  1. m_0 = \dfrac{M}{N_A}m_0 = \dfrac{0{,}046}{6\cdot10^{23}}m_0 = 7{,}7\cdot 10^{-26}\,\text{кг}
  2. p = \dfrac{1}{3}m_0 n\bar v^{2}p = \dfrac{1}{3}\cdot 7{,}7\cdot 10^{-26}\cdot 3\cdot 10^{25}\cdot 400^{2}p = 1{,}2\cdot 10^{5}\,\text{Па}

    Basic equation of molecular kinetic theory.

Answerp = 1{,}2\cdot 10^{5}\,\text{Па}

Position in this set: 6 from 24

Problem #76

Molecular kinetic theory: particles and their movement

Determine the average kinetic energy of a molecule and the root mean square velocity of oxygen and hydrogen molecules at 20\,^{\circ}\text{C}.

Answer options

  1. \left(1{,}22 \cdot 10^{-20}\,\text{Дж};\ 954\,\text{м/с};\ 3800\,\text{м/с}\right)
  2. \left(6{,}1 \cdot 10^{-20}\,\text{Дж};\ 4770\,\text{м/с};\ 19000\,\text{м/с}\right)
  3. \left(6{,}1 \cdot 10^{-21}\,\text{Дж};\ 477\,\text{м/с};\ 1900\,\text{м/с}\right)
  4. \left(3{,}05 \cdot 10^{-21}\,\text{Дж};\ 238{,}5\,\text{м/с};\ 950\,\text{м/с}\right)
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Given
t
20\,^{\circ}\text{C}T = 293\,\text{К}
M_1
32\cdot 10^{-3}\,\tfrac{\text{кг}}{\text{моль}}
M_2
2\cdot 10^{-3}\,\tfrac{\text{кг}}{\text{моль}}
k
1{,}38\cdot 10^{-23}\,\tfrac{\text{Дж}}{\text{К}}
R
8{,}31\,\tfrac{\text{Дж}}{\text{К}\cdot\text{моль}}
Find

\bar E\bar v_1\bar v_2

  1. \bar E = \dfrac{3}{2}kT\bar E = \dfrac{3}{2}\cdot 1{,}38\cdot 10^{-23}\cdot 293\bar E = 6{,}1\cdot 10^{-21}\,\text{Дж}

    The average energy depends only on temperature, so it is the same for oxygen and hydrogen.

  2. \bar v = \sqrt{\dfrac{3RT}{M}}\bar v = \sqrt{\dfrac{3\cdot 8{,}31\cdot 293}{32\cdot 10^{-3}}}\bar v_1 = 477\,\tfrac{\text{м}}{\text{с}}
  3. \bar v_2 = \bar v_1\sqrt{\dfrac{M_1}{M_2}} = 4\bar v_1\bar v_2 = 4\cdot 477\bar v_2 = 1908\,\tfrac{\text{м}}{\text{с}}

    Hydrogen is 16 times lighter than oxygen, so its molecules are 4 times faster.

Answer\bar E = 6{,}1\cdot 10^{-21}\,\text{Дж}\bar v_1 = 477\,\tfrac{\text{м}}{\text{с}}\bar v_2 = 1908\,\tfrac{\text{м}}{\text{с}}

Position in this set: 7 from 24

Problem #77

Molecular kinetic theory: gas laws

At temperature -15\,^{\circ}\text{C} the gas pressure in the vessel is equal to 1{,}2\cdot10^{5}\,\text{Па}. What will the pressure be at 15\,^{\circ}\text{C} at constant volume?

Answer options

  1. 67000\,\text{Па}
  2. 2{,}68 \cdot 10^{5}\,\text{Па}
  3. 1{,}34 \cdot 10^{6}\,\text{Па}
  4. 1{,}34 \cdot 10^{5}\,\text{Па}
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Given
t_1
-15\,^{\circ}\text{C}T_1 = 258\,\text{К}
t_2
15\,^{\circ}\text{C}T_2 = 288\,\text{К}
p_1
1{,}2\cdot 10^{5}\,\text{Па}
Find

p_2

  1. \dfrac{p_1}{T_1} = \dfrac{p_2}{T_2}

    The volume is constant - this is an isochoric process, Charles's law.

  2. p_2 = \dfrac{p_1 T_2}{T_1}p_2 = \dfrac{1{,}2\cdot 10^{5}\cdot 288}{258}p_2 = 1{,}34\cdot 10^{5}\,\text{Па}

Answerp_2 = 1{,}34\cdot 10^{5}\,\text{Па}

Position in this set: 8 from 24

Problem #78

Molecular kinetic theory: gas laws

In a cylinder with a volume 20\,\text{л} gas is under pressure 5\cdot10^{5}\,\text{Па} at 20\,^{\circ}\text{C}. How many molecules are in the gas and what volume does it occupy under normal conditions? Normal conditions: 273 K and 101325 Pa.

Answer options

  1. \left(2{,}47 \cdot 10^{24}\,\text{молекул};\ 91{,}96\,\text{л}\right)
  2. \left(1{,}235 \cdot 10^{24}\,\text{молекул};\ 45{,}98\,\text{л}\right)
  3. \left(4{,}94 \cdot 10^{24}\,\text{молекул};\ 183{,}92\,\text{л}\right)
  4. \left(2{,}47 \cdot 10^{25}\,\text{молекул};\ 919{,}6\,\text{л}\right)
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Given
V
20\,\text{л}20\cdot 10^{-3}\,\text{м}^{3}
p
5\cdot 10^{5}\,\text{Па}
T
293\,\text{К}
T_0
273\,\text{К}
p_0
101\,325\,\text{Па}
k
1{,}38\cdot 10^{-23}\,\tfrac{\text{Дж}}{\text{К}}
Find

NV_0

  1. p = nkT\ \Rightarrow\ N = nV = \dfrac{pV}{kT}N = \dfrac{5\cdot 10^{5}\cdot 20\cdot 10^{-3}}{1{,}38\cdot 10^{-23}\cdot 293}N = 2{,}47\cdot 10^{24}

    The Mendeleev-Clapeyron equation gives the same thing, because R = kN_A.

  2. \dfrac{pV}{T} = \dfrac{p_0 V_0}{T_0}\ \Rightarrow\ V_0 = \dfrac{pVT_0}{Tp_0}V_0 = \dfrac{5\cdot 10^{5}\cdot 20\cdot 10^{-3}\cdot 273}{293\cdot 101\,325}V_0 = 91{,}96\cdot10^{-3}\,\text{м}^{3}=91{,}96\,\text{л}

AnswerN = 2{,}47\cdot 10^{24}V_0=91{,}96\,\text{л}

Position in this set: 9 from 24

Problem #79

Molecular kinetic theory: gas laws

How many molecules are there in 5\,\text{м}^{3} gas at 27\,^{\circ}\text{C} and pressure 6\cdot10^{5}\,\text{Па}?

Answer options

  1. 7{,}25 \cdot 10^{27}\,\text{молекул}
  2. 7{,}25 \cdot 10^{26}\,\text{молекул}
  3. 3{,}625 \cdot 10^{26}\,\text{молекул}
  4. 1{,}45 \cdot 10^{27}\,\text{молекул}
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Given
p
6\cdot 10^{5}\,\text{Па}
V
5\,\text{м}^{3}
t
27\,^{\circ}\text{C}T = 300\,\text{К}
k
1{,}38\cdot 10^{-23}\,\tfrac{\text{Дж}}{\text{К}}
Find

N

  1. N = \dfrac{pV}{kT}N = \dfrac{6\cdot 10^{5}\cdot 5}{1{,}38\cdot 10^{-23}\cdot 300}N = 7{,}25\cdot 10^{26}

    The same technique as in problem 78.

AnswerN = 7{,}25\cdot 10^{26}

Position in this set: 10 from 24

Problem #80

Molecular kinetic theory: gas laws

What is the density of carbon dioxide under normal conditions? Take the molar mass of CO₂ to be 0.044 kg/mol; normal conditions: 273 K and 101325 Pa.

Answer options

  1. 3{,}94\,\text{кг/м}^{3}
  2. 19{,}7\,\text{кг/м}^{3}
  3. 1{,}97\,\text{кг/м}^{3}
  4. 0{,}985\,\text{кг/м}^{3}
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Given
\text{газ}
\mathrm{CO_2}
M
44\cdot 10^{-3}\,\tfrac{\text{кг}}{\text{моль}}
T_0
273\,\text{К}
p_0
101\,325\,\text{Па}
R
8{,}31\,\tfrac{\text{Дж}}{\text{К}\cdot\text{моль}}
Find

\rho_0

  1. pV = \dfrac{m}{M}RT\ \Rightarrow\ \rho = \dfrac{m}{V} = \dfrac{pM}{RT}

    We express density directly from the equation of state.

  2. \rho_0 = \dfrac{p_0 M}{RT_0}\rho_0 = \dfrac{101\,325\cdot 44\cdot 10^{-3}}{8{,}31\cdot 273}\rho_0 = 1{,}97\,\tfrac{\text{кг}}{\text{м}^{3}}

Answer\rho_0 = 1{,}97\,\tfrac{\text{кг}}{\text{м}^{3}}

Position in this set: 11 from 24

Problem #81

Molecular kinetic theory: particles and their movement

Available 30\,\text{моль} gas in volume 200\,\text{л} under pressure 8\cdot10^{5}\,\text{Па}. Determine the average kinetic energy of the molecule.

Answer options

  1. 6{,}65 \cdot 10^{-21}\,\text{Дж}
  2. 2{,}66 \cdot 10^{-20}\,\text{Дж}
  3. 1{,}33 \cdot 10^{-19}\,\text{Дж}
  4. 1{,}33 \cdot 10^{-20}\,\text{Дж}
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Given
\nu
30\,\text{моль}
V
200\,\text{л}0{,}2\,\text{м}^{3}
p
8\cdot 10^{5}\,\text{Па}
N_A
6\cdot 10^{23}\,\tfrac{1}{\text{моль}}
Find

\bar E

  1. pV = \nu RT\ \Rightarrow\ T = \dfrac{pV}{\nu R}
  2. \bar E = \dfrac{3}{2}kT = \dfrac{3kpV}{2\nu R} = \dfrac{3pV}{2\nu N_A}

    Boltzmann's constant cancels out because R = kN_A.

  3. \bar E = \dfrac{3pV}{2\nu N_A}\bar E = \dfrac{3\cdot 8\cdot 10^{5}\cdot 0{,}2}{2\cdot 30\cdot 6\cdot 10^{23}}\bar E = 1{,}33\cdot 10^{-20}\,\text{Дж}

Answer\bar E = 1{,}33\cdot 10^{-20}\,\text{Дж}

Position in this set: 12 from 24

Problem #92

Molecular kinetic theory: particles and their movement

The gas temperature changed from 127\,^{\circ}\text{C} up to 200\,\text{K}. How many times did the average kinetic energy and root mean square speed of the molecules change?

Answer options

  1. \left(1\,\text{раз};\ 1{,}4142\,\text{раз}\right)
  2. \left(5\,\text{раз};\ 7{,}071\,\text{раз}\right)
  3. \left(0{,}5\,\text{раз};\ 0{,}7071\,\text{раз}\right)
  4. \left(0{,}25\,\text{раз};\ 0{,}35355\,\text{раз}\right)
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Given
t_1
127\,^{\circ}\text{C}T_1 = 400\,\text{К}
T_2
200\,\text{К}
Find

\dfrac{\bar E_2}{\bar E_1}\dfrac{\bar v_2}{\bar v_1}

  1. \dfrac{T_1}{T_2} = \dfrac{400}{200} = 2

    The temperature dropped by half.

  2. \bar E = \dfrac{3}{2}kT \sim T\ \Rightarrow\ \dfrac{\bar E_2}{\bar E_1} = \dfrac{T_2}{T_1}\dfrac{\bar E_2}{\bar E_1} = 0{,}5

    Average energy is directly proportional to temperature.

  3. \bar v = \sqrt{\dfrac{3RT}{M}} \sim \sqrt{T}\ \Rightarrow\ \dfrac{\bar v_2}{\bar v_1} = \sqrt{\dfrac{T_2}{T_1}}\dfrac{\bar v_2}{\bar v_1} = \sqrt{0{,}5}\dfrac{\bar v_2}{\bar v_1} = 0{,}7071

Answer\dfrac{\bar E_2}{\bar E_1} = 0{,}5\dfrac{\bar v_2}{\bar v_1} = 0{,}7071

Position in this set: 13 from 24

Problem #183

Particles and amount of substance

Mass of substance 36 g, molar mass 18 g/mol. Find the amount of the substance.

Answer options

  1. 2\,\text{моль}
  2. 4\,\text{моль}
  3. 1\,\text{моль}
  4. 6\,\text{моль}
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Given
m
36\,\text{г}
M
18\,\text{г/моль}
Find

\nu

  1. \nu=\frac{m}{M}\nu=\frac{36}{18}\nu=2\,\text{моль}

    Mass and molar mass use the same unit of mass.

Answer2\,\text{моль}

Position in this set: 14 from 24

Problem #184

Particles and amount of substance

Mass of substance 88 g, molar mass 44 g/mol. Find the amount of the substance.

Answer options

  1. 4\,\text{моль}
  2. 2\,\text{моль}
  3. 1\,\text{моль}
  4. 6\,\text{моль}
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Given
m
88\,\text{г}
M
44\,\text{г/моль}
Find

\nu

  1. \nu=\frac{m}{M}\nu=\frac{88}{44}\nu=2\,\text{моль}

    Mass and molar mass use the same unit of mass.

Answer2\,\text{моль}

Position in this set: 15 from 24

Problem #185

Particles and amount of substance

Water mass 9 g has molar mass 18 g/mol. Find the number of molecules. Avogadro's constant Nₐ = 6{,}02\cdot 10^{23} mole⁻¹.

Answer options

  1. 6{,}02\cdot 10^{23}\,\text{молекул}
  2. 3{,}01\cdot 10^{23}\,\text{молекул}
  3. 1{,}505\cdot 10^{23}\,\text{молекул}
  4. 9{,}03\cdot 10^{23}\,\text{молекул}
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Given
m
9\,\text{г}
M
18\,\text{г/моль}
A
6{,}02\cdot 10^{23}\,\text{моль}^{-1}
Find

N

  1. \nu=\frac{m}{M}\nu=\frac{9}{18}\nu=0{,}5\,\text{моль}

    Determine the amount of substance.

  2. N=\nu N_AN=0{,}5 \cdot 6{,}02\cdot 10^{23}N=3{,}01\cdot 10^{23}\,\text{молекул}

    The number of particles is proportional to the amount of substance.

Answer3{,}01\cdot 10^{23}\,\text{молекул}

Position in this set: 16 from 24

Problem #186

Particles and amount of substance

Water mass 27 g has molar mass 18 g/mol. Find the number of molecules. Avogadro's constant Nₐ = 6{,}02\cdot 10^{23} mole⁻¹.

Answer options

  1. 1{,}806\cdot 10^{24}\,\text{молекул}
  2. 4{,}515\cdot 10^{23}\,\text{молекул}
  3. 9{,}03\cdot 10^{23}\,\text{молекул}
  4. 2{,}709\cdot 10^{24}\,\text{молекул}
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Given
m
27\,\text{г}
M
18\,\text{г/моль}
A
6{,}02\cdot 10^{23}\,\text{моль}^{-1}
Find

N

  1. \nu=\frac{m}{M}\nu=\frac{27}{18}\nu=1{,}5\,\text{моль}

    Determine the amount of substance.

  2. N=\nu N_AN=1{,}5 \cdot 6{,}02\cdot 10^{23}N=9{,}03\cdot 10^{23}\,\text{молекул}

    The number of particles is proportional to the amount of substance.

Answer9{,}03\cdot 10^{23}\,\text{молекул}

Position in this set: 17 from 24

Problem #187

Particles and amount of substance

In a vessel with a volume 10 l is located 4 g of gas with molar mass 4 g/mol. Find the concentration of molecules in m⁻³, Nₐ = 6{,}02\cdot 10^{23} mole⁻¹.

Answer options

  1. 1{,}204\cdot 10^{26}\,\text{м}^{-3}
  2. 3{,}01\cdot 10^{25}\,\text{м}^{-3}
  3. 6{,}02\cdot 10^{25}\,\text{м}^{-3}
  4. 1{,}806\cdot 10^{26}\,\text{м}^{-3}
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Given
V
10\,\text{л}
m
4\,\text{г}
M
4\,\text{г/моль}
A
6{,}02\cdot 10^{23}\,\text{моль}^{-1}
Find

n

  1. v=\frac{V}{1000}v=\frac{10}{1000}v=0{,}01\,\text{м}^{3}

    We reduce the volume to SI.

  2. N=\frac{m}{M}N_AN=\frac{4}{4} \cdot 6{,}02\cdot 10^{23}N=6{,}02\cdot 10^{23}\,\text{молекул}

    Find the number of molecules in the vessel.

  3. n=\frac{N}{v}n=\frac{6{,}02\cdot 10^{23}}{0{,}01}n=6{,}02\cdot 10^{25}\,\text{м}^{-3}

    Concentration is the number of particles per unit volume.

Answer6{,}02\cdot 10^{25}\,\text{м}^{-3}

Position in this set: 18 from 24

Problem #188

Particles and amount of substance

In a vessel with a volume 20 l is located 12 g of gas with molar mass 4 g/mol. Find the concentration of molecules in m⁻³, Nₐ = 6{,}02\cdot 10^{23} mole⁻¹.

Answer options

  1. 1{,}806\cdot 10^{26}\,\text{м}^{-3}
  2. 4{,}515\cdot 10^{25}\,\text{м}^{-3}
  3. 2{,}709\cdot 10^{26}\,\text{м}^{-3}
  4. 9{,}03\cdot 10^{25}\,\text{м}^{-3}
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Given
V
20\,\text{л}
m
12\,\text{г}
M
4\,\text{г/моль}
A
6{,}02\cdot 10^{23}\,\text{моль}^{-1}
Find

n

  1. v=\frac{V}{1000}v=\frac{20}{1000}v=0{,}02\,\text{м}^{3}

    We reduce the volume to SI.

  2. N=\frac{m}{M}N_AN=\frac{12}{4} \cdot 6{,}02\cdot 10^{23}N=1{,}806\cdot 10^{24}\,\text{молекул}

    Find the number of molecules in the vessel.

  3. n=\frac{N}{v}n=\frac{1{,}806\cdot 10^{24}}{0{,}02}n=9{,}03\cdot 10^{25}\,\text{м}^{-3}

    Concentration is the number of particles per unit volume.

Answer9{,}03\cdot 10^{25}\,\text{м}^{-3}

Position in this set: 19 from 24

Problem #189

Ideal gases and constant-parameter processes

An ideal gas at constant temperature is compressed with 6 up to 2 l. Initial pressure 100 kPa. Find the final pressure.

Answer options

  1. 600\,\text{кПа}
  2. 300\,\text{кПа}
  3. 150\,\text{кПа}
  4. 900\,\text{кПа}
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Given
V_1
6\,\text{л}
V_2
2\,\text{л}
p_1
100\,\text{кПа}
Find

p_2

  1. p_2=p_1\frac{V_1}{V_2}p_2=\frac{100 \cdot 6}{2}p_2=300\,\text{кПа}

    At constant mass and temperature, the product pV is conserved.

Answer300\,\text{кПа}

Position in this set: 20 from 24

Problem #190

Ideal gases and constant-parameter processes

An ideal gas at constant temperature is compressed with 8 up to 4 l. Initial pressure 150 kPa. Find the final pressure.

Answer options

  1. 600\,\text{кПа}
  2. 150\,\text{кПа}
  3. 300\,\text{кПа}
  4. 900\,\text{кПа}
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Given
V_1
8\,\text{л}
V_2
4\,\text{л}
p_1
150\,\text{кПа}
Find

p_2

  1. p_2=p_1\frac{V_1}{V_2}p_2=\frac{150 \cdot 8}{4}p_2=300\,\text{кПа}

    At constant mass and temperature, the product pV is conserved.

Answer300\,\text{кПа}

Position in this set: 21 from 24

Problem #191

Ideal gases and constant-parameter processes

A gas in a rigid closed vessel is heated from 27 up to 127 °C. Initial pressure 150 kPa. Find the final pressure, taking T(K) = t(°C) + 273.

Answer options

  1. 400\,\text{кПа}
  2. 100\,\text{кПа}
  3. 200\,\text{кПа}
  4. 600\,\text{кПа}
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Given
T_1
27\,\text{°C}
T_2
127\,\text{°C}
p_1
150\,\text{кПа}
Find

p_2

  1. T_a=t_1+273T_a=27 + 273T_a=300\,\text{К}

    We use absolute temperature.

  2. T_b=t_2+273T_b=127 + 273T_b=400\,\text{К}

    Convert the final temperature.

  3. p_2=p_1\frac{T_b}{T_a}p_2=\frac{150 \cdot 400}{300}p_2=200\,\text{кПа}

    At constant volume, pressure is proportional to T.

Answer200\,\text{кПа}

Position in this set: 22 from 24

Problem #192

Ideal gases and constant-parameter processes

A gas in a rigid closed vessel is heated from 27 up to 177 °C. Initial pressure 200 kPa. Find the final pressure, taking T(K) = t(°C) + 273.

Answer options

  1. 600\,\text{кПа}
  2. 150\,\text{кПа}
  3. 900\,\text{кПа}
  4. 300\,\text{кПа}
Show the walkthroughHide the walkthrough
Given
T_1
27\,\text{°C}
T_2
177\,\text{°C}
p_1
200\,\text{кПа}
Find

p_2

  1. T_a=t_1+273T_a=27 + 273T_a=300\,\text{К}

    We use absolute temperature.

  2. T_b=t_2+273T_b=177 + 273T_b=450\,\text{К}

    Convert the final temperature.

  3. p_2=p_1\frac{T_b}{T_a}p_2=\frac{200 \cdot 450}{300}p_2=300\,\text{кПа}

    At constant volume, pressure is proportional to T.

Answer300\,\text{кПа}

Position in this set: 23 from 24

Problem #193

Ideal gases and constant-parameter processes

Gas volume 6 l at pressure 100 kPa and temperature 300 K is transferred to the state 4 l, 400 K. Find the final pressure at constant mass.

Answer options

  1. 400\,\text{кПа}
  2. 100\,\text{кПа}
  3. 600\,\text{кПа}
  4. 200\,\text{кПа}
Show the walkthroughHide the walkthrough
Given
V_1
6\,\text{л}
p_1
100\,\text{кПа}
T_1
300\,\text{К}
V_2
4\,\text{л}
T_2
400\,\text{К}
Find

p_2

  1. a=\frac{V_1}{V_2}a=\frac{6}{4}a=1{,}5\,\text{раз}

    Decreasing volume increases pressure.

  2. b=\frac{T_2}{T_1}b=\frac{400}{300}b=1{,}33333\,\text{раз}

    An increase in absolute temperature increases pressure.

  3. p_2=p_1\frac{V_1T_2}{V_2T_1}p_2=100 \cdot 1{,}5 \cdot 1{,}33333p_2=200\,\text{кПа}

    We apply the unified gas law.

Answer200\,\text{кПа}

Position in this set: 24 from 24

Problem #194

Ideal gases and constant-parameter processes

Gas volume 8 l at pressure 150 kPa and temperature 300 K is transferred to the state 6 l, 450 K. Find the final pressure at constant mass.

Answer options

  1. 300\,\text{кПа}
  2. 600\,\text{кПа}
  3. 150\,\text{кПа}
  4. 900\,\text{кПа}
Show the walkthroughHide the walkthrough
Given
V_1
8\,\text{л}
p_1
150\,\text{кПа}
T_1
300\,\text{К}
V_2
6\,\text{л}
T_2
450\,\text{К}
Find

p_2

  1. a=\frac{V_1}{V_2}a=\frac{8}{6}a=1{,}33333\,\text{раз}

    Decreasing volume increases pressure.

  2. b=\frac{T_2}{T_1}b=\frac{450}{300}b=1{,}5\,\text{раз}

    An increase in absolute temperature increases pressure.

  3. p_2=p_1\frac{V_1T_2}{V_2T_1}p_2=150 \cdot 1{,}33333 \cdot 1{,}5p_2=300\,\text{кПа}

    We apply the unified gas law.

Answer300\,\text{кПа}

The theory behind these problems

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Particles and amount of substanceMolecular mass, the mole, particle count and number density.Ideal gases and constant-parameter processesGas pressure, temperature, equation of state and gas processes.

Formulas by topic

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Amount of substanceν · мольMolecular concentrationn · м⁻³Molar massM · кг/мольVelocityv · м/сTemperatureT, t · К, °C