A gas is stored in an isochoric, refrigerated tank that has V = 5 m³. Initially, the gas inside the tank has T = 15°C and P = 5 bar, while the ambient surroundings are at 25°C and atmospheric pressure. The refrigeration system fails, and the gas inside the tank gradually warms to 25°C. A) Find

Question

A gas is stored in an isochoric, refrigerated tank that has V = 5 m³ . Initially, the gas inside the tank has T = 15°C and P = 5 bar, while the ambient surroundings are at 25°C and atmospheric pressure. The refrigeration system fails, and the gas inside the tank gradually warms to 25°C.

A) Find the final pressure of the gas, assuming it is an ideal gas.
B) Find the final pressure of the gas, assuming the gas is described by the van der Waals equation of state, with [latex]a=8.0 imes { 10 }^{ 5 } m{ cm }^{ 6 }/ m bar/{ m mol }^{ 2 }[/latex] and b = 100 [latex] m{ cm^3 }/{ m mol }[/latex].
C) For the cases in parts A and B, how much work was done by the gas on the surroundings?

Accepted Answer

In this problem, we will use the Ideal Gas Constant [latex]\left(0.08314 \frac{\mathrm{L} \,\mathrm{bar}}{\mathrm{mol} \,\mathrm{K}} ight)\left(\frac{1 \mathrm{~m}^{3}}{1000 \mathrm{~L}} ight)=8.314 imes10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}}[/latex]

A) [latex]25^{\circ} \mathrm{C}=298 \mathrm{~K} \quad 15^{\circ} \mathrm{C}=288 \mathrm{~K}[/latex]

[latex]\mathrm{PV}=\mathrm{NRT}[/latex]

[latex]\frac{\mathrm{T}}{\mathrm{P}}=\frac{\mathrm{V}}{\mathrm{NR}}[/latex]

With N, R and V all constant:

[latex]\begin{aligned}& \frac{\mathrm{T}_{ ext {final }}}{\mathrm{P}_{ ext {final }}}=\frac{\mathrm{T}_{ ext {initial }}}{\mathrm{P}_{ ext {initial }}}\& \mathrm{P}_{ ext {final }}=\frac{\mathrm{T}_{ ext {final }} \mathrm{P}_{\mathrm{initial}}}{\mathrm{T}_{ ext {initial }}}=\frac{(298 \mathrm{~K})(5 \mathrm{bar})}{(288 \mathrm{~K})}=\mathbf{5 . 1 7 \mathbf { b a r }}\end{aligned}[/latex]

B) Find the molar volume of gas at the initial temperature and pressure, using the VDW equation:

[latex]\begin{aligned}& \mathrm{P}=\frac{\mathrm{RT}}{\underline{\mathrm{V}}-\mathrm{b}}-\frac{\mathrm{a}}{\underline{\mathrm{V}}^{2}}\& 5 \mathrm{bar}=\frac{\left(8.314 imes 10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \mathrm{K}} ight) 288 \mathrm{~K}}{(\underline{\mathrm{V}})-\left(\frac{100 \mathrm{~cm}^{3}}{\mathrm{~mol}} ight)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}} ight)^{3}}-\frac{\left(8.0 imes 10^{6} \frac{\mathrm{cm}^{6} \mathrm{bar}}{\mathrm{mol}^{2}} ight)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}} ight)^{6}}{(\underline{\mathrm{V}})^{2}}\end{aligned}[/latex]

Solving this cubic equation for " V " yields three answers, only one of which is real. Therefore,

[latex]\underline{V}=0.004544 \frac{\mathrm{m}^{3}}{\mathrm{~mol}}[/latex]

Solving for N using known total volume:

[latex]\begin{aligned}& V=N \underline{V}\& N=\frac{5 \mathrm{~m}^{3}}{0.004544 \frac{\mathrm{m}^{3}}{\mathrm{~mol}}}=1100 . \mathrm{mol}\end{aligned}[/latex]

Since system is closed and isochoric, N, V and V are all constant. At final conditions:

[latex]\begin{aligned}& P=\frac{R T}{\underline{V}-b}-\frac{a}{\underline{V}^{2}}\& P=\frac{\left(8.314 imes 10^{-5} \frac{\mathrm{m}^{3} \mathrm{bar}}{\mathrm{mol} \,\mathrm{K}} ight) 298 \mathrm{~K}}{\left(\frac{5 \mathrm{~m}^{3}}{1100 \mathrm{mols}} ight)-\left(\frac{100 \mathrm{~cm}^{3}}{\mathrm{~mol}} ight)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}} ight)^{3}}-\frac{\left(8.0 imes 10^{6} \frac{\mathrm{cm}^{6} \mathrm{bar}}{\mathrm{mol}^{2}} ight)\left(\frac{\mathrm{m}}{100 \mathrm{~cm}} ight)^{6}}{\left(\frac{5 \mathrm{~m}^{3}}{1100 \mathrm{mols}} ight)^{2}}\& \qquad =\bf 5.19 \,\mathbf{bar}\end{aligned}[/latex]

C) No work was done on the gas by the environment. Heat was added, but there was no expansion or contraction, and therefore no [latex]\mathrm{W}_{\mathrm{EC}}[/latex].

Question 2.20: A gas is stored in an isochoric, refrigerated tank that has ...

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