Solved thermodynamics exercise

Heating at constant pressure: deriving cPc_P

Exercise 13 · Lesson 5 — Application to the Ideal Gas

  • heat capacity
  • constant pressure
  • Mayer's relation
  • isobaric heating
  • ideal gas

Statement

One mole of an ideal gas with molar heat capacity cVc_V is enclosed in a cylinder with initial volume V0V_0. The cylinder is closed by a frictionless movable adiabatic piston subjected to a constant external pressure P0P_0. An electrical resistor inside the gas supplies it with heat Q>0Q>0 quasi-statically.

  1. Draw a simple diagram of the apparatus. Indicate the gas, the piston, the resistor, the external pressure P0P_0, and the energy transfers.
  2. Without using numerical values, express the initial temperature T0T_0, then determine the final state (Pf,Vf,Tf)(P_f,V_f,T_f), the work WW received by the gas, and its change in internal energy ΔU\Delta U.
  3. Deduce the molar heat capacity at constant pressure cP=(δQ/dT)Pc_P=(\delta Q/\mathrm{d}T)_P in terms of cVc_V and RR.
  4. Numerical application for a monatomic ideal gas: cV=32Rc_V=\frac32R, V0=25 LV_0=25\ \mathrm{L}, P0=1,0×105 PaP_0=1{,}0\times10^5\ \mathrm{Pa}, Q=1000 JQ=1000\ \mathrm{J}, and R=8,314 J mol−1 K−1R=8{,}314\ \mathrm{J\,mol^{-1}\,K^{-1}}.

Hint
The transformation is isobaric: W=−P0 ΔVW=-P_0\,\Delta V. For one mole of an ideal gas, P0 ΔV=R ΔTP_0\,\Delta V=R\,\Delta T. Substitute these relations into the first law.

Detailed solution

Solution
Question 1. The diagram should show the gas in a cylinder closed by a movable piston. An arrow inside the cylinder represents the heat QQ supplied by the resistor. The external pressure P0P_0 acts on the piston, which moves upward when the gas expands.

Question 2. The initial equation of state gives

T0=P0V0R.T_0=\frac{P_0V_0}{R}.

The pressure remains constant, so Pf=P0P_f=P_0. Moreover, W=−P0ΔV=−RΔTW=-P_0\Delta V=-R\Delta T. The first law then gives

ΔU=cVΔT=Q+W=Q−RΔT,\Delta U=c_V\Delta T=Q+W=Q-R\Delta T,

hence

ΔT=QcV+R.\Delta T=\frac{Q}{c_V+R}.

The final state is therefore

Pf=P0,Tf=T0+QcV+R,Vf=V0+RQP0(cV+R).P_f=P_0,\qquad T_f=T_0+\frac{Q}{c_V+R},\qquad V_f=V_0+\frac{RQ}{P_0(c_V+R)}.

Finally,

W=−RQcV+R,ΔU=cVQcV+R.W=-\frac{RQ}{c_V+R}, \qquad \Delta U=\frac{c_VQ}{c_V+R}.

Question 3. Since Q=(cV+R)ΔTQ=(c_V+R)\Delta T at constant pressure,

cP=cV+R.c_P=c_V+R.

This is Mayer's relation.

Question 4. For a monatomic gas,

cP=52R≃20,8 J mol−1 K−1.c_P=\frac52R\simeq20{,}8\ \mathrm{J\,mol^{-1}\,K^{-1}}.

We obtain

T0≃301 K,ΔT≃48,1 K,Tf≃349 K,Vf=29,0 L.T_0\simeq301\ \mathrm{K},\qquad \Delta T\simeq48{,}1\ \mathrm{K},\qquad T_f\simeq349\ \mathrm{K},\qquad V_f=29{,}0\ \mathrm{L}.

Finally,

W=−400 J,ΔU=600 J.W=-400\ \mathrm{J},\qquad \Delta U=600\ \mathrm{J}.