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.
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problems.proof.prevStepproblems.proof.nextStep
PV=RgT,PΔV=Rg(T2−T1)PV=R_gT,\quad P\Delta V=R_g(T_2-T_1)
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For one mole at constant pressure, the equation of state gives PΔV=Rg(T2−T1)P\Delta V=R_g(T_2-T_1). Pressure remains constant and the gas follows the ideal-gas model.

problems.proof.pitfall. For a monatomic gas, CV=32RgC_V=\tfrac32R_g; also, since WW is work done by the gas, Q=ΔU+WQ=\Delta U+W, not ΔU−W\Delta U-W.