Internal energy and the first law

Gas in a thermally insulated vessel lifts a piston and does work. Where did the energy transferred out come from if no heat entered? To answer, distinguish the energy stored in the gas from the ways energy crosses its boundary.

You can carry a sealed thermos quickly without heating its contents. Motion of the whole system must be distinguished from motion and interactions of its particles. The latter determine internal energy U, measured in joules. Choosing the gas as our system, we compare its final and initial states: \Delta U=U_2-U_1. A positive difference means its internal energy increased; a negative difference means it decreased. The difference alone does not tell us how energy was transferred.

Consider a monatomic ideal gas: we neglect interactions between collisions and need not include molecular rotation or vibration. Each particle has average translational energy 3kT/2, so summing over the particles gives U=3\nu RT/2. Here \nu=m/M is in moles and T in K. Doubling the amount of gas at the same temperature doubles U. The factor 3/2 comes from the three directions of translational motion.

If pressure and volume are known instead of temperature, no new particle model is needed. Using pV=\nu RT gives U=3mRT/(2M)=3pV/2 for the same monatomic ideal gas. Pa·m³ gives joules. For a fixed amount of gas, equal products pV mean equal U, even when p and V differ individually. A polyatomic gas may have additional rotational and vibrational contributions; the factor 3/2 cannot automatically be used for its total internal energy.

To find the change in stored energy, subtract 3mRT/(2M) for two states of the same sample: \Delta U=3mR\Delta T/(2M). Here \Delta T=T_2-T_1; cooling gives a negative temperature change and a negative ΔU. At the same mass, doubling the temperature change doubles the energy change. Use consistent units for m and M: kg and kg/mol. A temperature difference has the same numerical value in K and °C, although temperatures in the formula for U must be in K.

When gas expands and pushes a piston, it transfers energy to its surroundings. Call the work done by the gas A, positive for expansion. During compression, the surroundings transfer energy to the gas: their work A\prime is positive and A is negative. For the same transfer, A\prime=-A. The sign does not measure how useful the process is; it records which side of the system boundary we use to describe the energy transfer.

Let the gas slowly raise a piston of area S through distance l at constant pressure p. The force is pS, so the work is A=pSl=p\Delta V, since \Delta V=Sl=V_2-V_1. Doubling the volume increase doubles the work; compression gives negative ΔV. Pressure in Pa and volume in m³ give J. Slow motion lets us treat the gas as close to equilibrium; during a rapid process, pressure may differ across the gas.

If piston travel is hard to measure, use temperature. For a fixed amount of ideal gas at constant p, subtracting the equations of state gives p\Delta V=(m/M)R\Delta T. Hence A=(m/M)R\Delta T: on heating, the gas expands and does positive work. If pressure varies, multiplying one pressure by the total volume change is insufficient; calculate work over small intervals while accounting for the changing pressure.

Gas can receive energy by heat transfer and give energy out by doing work. Define received heat Q as positive and released heat as negative. Conservation of energy then gives \Delta U=Q-A, or Q=\Delta U+A: incoming energy either changes internal energy or leaves through work. When Q = 0, positive work reduces U. Heat is not a substance stored inside: Q describes a transfer, whereas U describes the system's state.

A problem may give the work done on the gas. Then write \Delta U=Q+A\prime: both positive terms describe incoming energy. If the gas is compressed without heat transfer, \Delta U=A\prime>0 and its internal energy increases. This is the same first law, because A\prime=-A. Before calculating, label the work in words: 'done by the gas' or 'done on the gas'. This prevents changing a sign without a physical reason.

Heat gas in a rigid vessel. Its boundary does not move, so expansion work A = 0. If there are no other types of work, the first law gives Q=\Delta U: all energy transferred by heat increases internal energy; during cooling, both quantities are negative. Constant volume alone does not prohibit all work: a stirrer could transfer energy to the gas at fixed V. Our constant-volume example excludes such work.

Expand the same amount of ideal gas at constant temperature. For an ideal gas, U depends on temperature and amount, so \Delta U=0. Yet the piston moves and work can be positive, requiring Q=A. Incoming heat replaces energy transferred out as work. Thus 'constant temperature' does not mean 'no heat transfer'. During isothermal compression, work and heat change sign: energy must be removed.

Under a constant piston load, heating a monatomic ideal gas has two effects: it increases the thermal motion of particles and expands the gas. The first contribution is \Delta U=3\nu R\Delta T/2, the second is A=\nu R\Delta T, and together Q=\Delta U+A. The same temperature rise therefore requires more heat at constant pressure than at constant volume: some energy leaves through the moving piston. The process conditions determine how the energy is shared.

Adding the contributions gives Q=5\nu R\Delta T/2. Internal energy receives 3/5 of this amount and work accounts for 2/5: \Delta U=0{,}6Q, A=0{,}4Q. If the heat input doubles under the same conditions, both contributions double but their shares stay the same. This follows from the specific model of a monatomic ideal gas at constant pressure with only expansion work. For another gas or a fixed piston, these shares do not apply.

A gas received 300 J of heat and did 120 J of work. How did its internal energy change?

Q=+300 J: energy entered as heat. A=+120 J: the gas transferred energy out as work.

Choose ΔU=Q−A because A is work done by the gas.

ΔU=300−120=180 J. The positive sign means internal energy increased.

Check: 180 J remained in the gas and 120 J left as work; together these equal the 300 J received.

The internal energy increased by 180 J.

Internal energy is stored energy associated with the state; heat and work change it by transferring energy across the system boundary.

The same balance can be written ΔU = Q − A or ΔU = Q + A′; the sign depends on who does the work.

The formula U = 3νRT/2 and the shares 0.6Q/0.4Q require a monatomic ideal gas and the stated process conditions.

Next, we examine heating, phase changes, and heat engines.