Question

One mole of an ideal gas in a closed system, initially at 25°C and 10 bar, is first expanded adiabatically, then heated isochorically to reach a final state of 25°C and 1 bar. Assuming these processes are mechanically reversible, compute T and P after the adiabatic expansion, and compute Q, W, ΔU, and ΔH for each step and for the overall process. Take CP = (7/2)R and CV = (5/2)R.

Answer 1

Answer:

$P_2=0.398bar=39800Pa$

$T_2=118.7K$

$Q=-3729.9J$

$W=-61753.24J$

Δ$U_T=0J$

Δ$H_T=0J$

Explanation:

Hello,

At the first state, the molar volume is:

$v_1=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^6Pa}=2.48x10^{-3}m^3$

The volume in both the second and third state:

$v_2=v_3=\frac{RT}{P_1} =\frac{8.314\frac{Pa*m^3}{molK}*298.15}{1x10^5Pa}=2.48x10^{-2}m^3$

Now, as it is about an adiabatic process, one remembers the following relationships:

$PV^\alpha =K\\TV^{\alpha-1}\\\alpha=\frac{Cp}{Cv}=\frac{7/2R}{5/2R}=1.4$

- Next, for the aforesaid volumes and the first pressure, one computes the second pressure as:

$P_2=\frac{P_1V_1^\alpha }{V_2^\alpha} =\frac{10bar*(2.48x10^{-3}m^3)^{1.4}}{(2.48x10^{-2}m^3)^{1.4}} =0.398bar=39800Pa$

- And the temperature:

$T_2=\frac{T_1V_1^{\alpha-1}}{V_2^{\alpha-1}} =\frac{298.15K*(2.48x10^{-3}m^3)^{1.4-1}}{(2.48x10^{-2}m^3)^{1.4-1}} =118.7K$

- Q:

It is clear that the heat for the first process is 0 as it is adiabatic, but for the second one, it is computed as:

$Q_2=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J$

Then the total heat:

$Q=Q_1+Q_2=0-3729.9J=-3729.9J$

- The work for the first process is:

$W_1=\frac{P_2V_2-P_1V_1}{1-\alpha }=\frac{39800Pa*2.48x10^{-3}m^3-1x10^6Pa*2.48x10^{-2}m^3}{0.4} \\W_1=-61753.24J$

It is clear that the second process is isochoric, so the work here is zero, thus, the total work is:

$W=W_1+W_2=-61753.24J+0J=-61753.24J$

- For the two processes, ΔU becomes the same value since the system returns to the initial temperature, so ΔU total is 0, thus, for each process, one's got:

$U_1=nCv(T_2-T_1)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-3729.9J\\U_2=nCv(T_3-T_2)=1mol*\frac{5}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=3729.9J$

- Finally, the total enthapy is also 0 due to same aforesaid reason, thus, each enthalpy is:

$H_1=nCp(T_2-T_1)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(118.7K-298.15K)=-5221.86J\\H_2=nCv(T_3-T_2)=1mol*\frac{7}{2}(8.314\frac{J}{mol*K})*(298.15K-118.7K)=5221.86J$

Best regards.