Two moles of an ideal gas (Cp=7/2 R) is allowed to expand adiabatically and reversibly at 27 °C from a pressure of 10 atm to a pressure of 1 atm. Calculate the final temperature of the gas and the amount of work done by the gas.

Question

Two moles of an ideal gas [latex](C_p=7/2 R)[/latex] is allowed to expand adiabatically and reversibly at 27 °C from a pressure of 10 atm to a pressure of 1 atm. Calculate the final temperature of the gas and the amount of work done by the gas.

Accepted Answer

[latex]\begin{array}{l} ext { Since } \mathrm{C}_{\mathrm{P}}=\frac{7}{2} \mathrm{R} \ \\mathrm{C}_{\mathrm{V}}=\mathrm{C}_{\mathrm{P}}-\mathrm{R}=\frac{7}{2} \mathrm{R}-\mathrm{R}=\frac{5}{2} \mathrm{R}\ \ herefore \gamma=\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}}=\frac{\frac{7}{2} \mathrm{R}}{\frac{5}{2} \mathrm{R}}=\frac{7}{5}=1.4\end{array}[/latex]

For an adiabatic expansion of an ideal gas

[latex]\left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}} ight)^{\gamma}=\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}} ight)^{\gamma-1}[/latex]

Here [latex]\mathrm{P}_{1}=10 \mathrm{~atm} ; \quad \mathrm{P}_{2}=1 \mathrm{~atm} ;\quad \mathrm{T}_{1}=273+27=300 \mathrm{~K} \quad \gamma=1.4[/latex]

[latex]\begin{array}{c} herefore\left(\frac{\mathrm{T}_{2}}{300} ight)^{1.4}=\left(\frac{1}{10} ight)^{1.4-1} \ \ \begin{aligned} ext { or } \mathrm{T}_{2}&=300 imes\left(\frac{1}{10} ight)^{\frac{0.4}{1.4}}=155.38 \mathrm{~K} \&=155.3-273=-117.61 ^{\circ} \mathrm{C}\end{aligned}\end{array}[/latex]

Therefore, final temperature of the gas [latex]T_2= –117.61 °C[/latex].

The work done by the gas in a reversible adiabatic expansion is given by

[latex]\begin{array}{l}\mathrm{w}=\mathrm{nC}_{\mathrm{v}} \mathrm{T}_{1}\left(\left(\frac{\mathrm{P}_{2}}{\mathrm{P}_{1}} ight)^{\frac{\gamma-1}{\gamma}}-1 ight) \ \\mathrm{n}=2, \quad \mathrm{C}_{\mathrm{v}}=\frac{5}{2} \mathrm{R}, \mathrm{T}_{1}=300 \mathrm{~K} \quad \gamma=1.4 \quad \mathrm{P}_{2}=1 \mathrm{~atm} \quad \mathrm{P}_{1}=10 \mathrm{~atm} \ \\mathrm{w}=2 imes \frac{5}{2} imes 1.987 imes 300\left(\left(\frac{1}{10} ight)^{\frac{1.4-1}{1.4}}-1 ight)=-1436.79 \mathrm{~cal}\end{array}[/latex]

Negative sign indicates the work expansion.

Question 15.2.2: Two moles of an ideal gas (Cp=7/2 R) is allowed to expand ad......

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