Physic Labs

Problem 2

One mole of a monatomic ideal gas, initially at absolute temperature T1T_1, is heated to T2>T1T_2>T_1 in a constant-pressure process. Let RgR_g be the gas constant; assume thermodynamic equilibrium throughout and no phase change. Define WW as work done by the gas and Q>0Q>0 as heat absorbed, so the first law is ΔU=Q−W\Delta U=Q-W. (a) Use the equation of state to find the volume change and work. (b) Calculate the internal-energy change from the molecular degrees of freedom. (c) Determine the heat transfer, check its sign, and explain how it is divided between raising internal energy and doing work.
Q=ΔU+W=52Rg(T2−T1)Q=\Delta U+W=\frac52R_g(T_2-T_1)
problems.proof.analysis

With work defined as done by the gas, the first law gives Q=ΔU+W=5RgΔT/2Q=\Delta U+W=5R_g\Delta T/2. Pressure remains constant and the gas follows the ideal-gas model.