Question

A sample of $3.0$ moles of perfect gas at $200$ K and $2.0$ atm is compressed reversibly and adiabatically until the temperature reaches $250$ K. Given that molar heat capacity is $27.5$ J K${}^{-1}$ mol${}^{-1}$ at constant volume. Calculate $q,w,$ $\Delta U$, $\Delta H$ and the final pressure and volume.

• A. $P=6.28$ atm, $V=11.75$ litre, $w=5.158$ kJ, $q=$ 0, $\Delta U=5.157$ kJ, $\Delta H=5.672$ kJ
• B. $P=5.28$ atm, $V=11.85$ litre, $w=4.158$ kJ, $q=$ 0, $\Delta U=5.157$ kJ, $\Delta H=5.672$ kJ
• C. $P=5.26$ atm, $V=11.81$ litre, $w=4.157$ kJ, $q=$ 0, $\Delta U=4.157$ kJ, $\Delta H=5.372$ kJ
• D. $P=5.26$ atm, $V=11.82$ litre, $w=4.152$ kJ, $q=$ 0, $\Delta U=4.157$ kJ, $\Delta H=5.372$ kJ

Answer 1

Answer: (C)

$P=5.26$ atm, $V=11.81$ litre, $w=4.157$ kJ, $q=$ 0, $\Delta U=4.157$ kJ, $\Delta H=5.372$ kJ

The initial values are $n=3mole,T_1=200K,P_1=2.0atm,C_v=27.5J{K}^{-1}mo{l}^{-1}$

The final values (after compression) are $T_2=250K,P_2=?$

The heat capacity at constant pressure is $C_p=27.5+8.314J{K}^{-1}mo{l}^{-1}=35.814J{K}^{-1}mo{l}^{-1}$

Also $\gamma =\frac{C_p}{C_v}=\frac{35.814}{27.5}=1.30$

But $p_1^{1-\gamma }\cdot T_1^{\gamma }=P_2^{1-\gamma }\cdot T_2^{\gamma }$

or $P_2^{1-\gamma }=P_1^{1-\gamma }{\left(\frac{T_1}{T_2}\right)}^{\gamma }$

Substitute values in the above expression.

$(P_2{)}^{-0.3}=(2{)}^{-0.3}\times {\left(\frac{200}{250}\right)}^{1.30}$

Thus, $P_2=5.26atm$

The initial volume is $V_1=\frac{nRT}{P_1}=\frac{3\times 0.0821\times 200}{2}=24.63litre$

Also, $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

or ${\left[\frac{V_2}{V_1}\right]}^{\gamma }=\frac{P_1}{P_2}$

Substitute values in the above expression.

${\left[\frac{V_2}{24.63}\right]}^{1.3}=\frac{2}{5.2}$

Hence, $V_2=11.81litre$

The work done is $=\frac{nR}{\gamma -1}[T_2-T_1]=\frac{3\times 8.314}{0.3}\times [250-200]=+4157J=+4.157kJ$

Since the process is adiabatic, no heat is transferred. $q=0$

Internal energy decreases as work is done. Hence, $\Delta U=w=4.157kJ$

Also, the expression for the enthalpy change is $\Delta H=n\times C_p\times \Delta T$

Substitute values in the above expression.

$\Delta H=3\times 35.814\times 50=5372.1J=5.372kJ$