Question

During an experiment, an ideal gas is found to obey an additional law pV² = constant. The gas is initially at a temperature T and volume V. Find the temperature when it expands to a volume 2V.

Answer 1

$\begin{array}{l}\text{Applying equation of state of an ideal gas, we get}\\ PV=nRT\\ \Rightarrow P=\frac{nRT}{V}...1\\ \text{Taking differentials, we get}\\ \Rightarrow PdV+VdP=nRdT...2\\ \text{Applying the additional law, we get}\\ P{V}^2=c\\ V^{2}dP+2VPdV=0\\ \Rightarrow VdP+2PdV=0...3\\ \text{Subtracting eq.}\left(\text{3}\right)\text{from eq.}\left(\text{2}\right)\text{, we get}\\ PdV=-nRdT\\ \Rightarrow dV=-\frac{nR}{P}dT\\ \text{Now,}\\ \text{⇒}dV=-\frac{V}{T}dT\left[\mathrm{From}\mathrm{eq}.\left(1\right)\right]\\ \Rightarrow \frac{dV}{V}=-\frac{dT}{T}\\ {\text{Integrating between T}}_2{\text{and T}}_1\text{, we get}\\ \Rightarrow \underset{V_1}{\overset{2V}{\int }}=-\underset{T_1}{\overset{T_2}{\int }}\\ \Rightarrow \ln \left(2V\right)-ln\left(V\right)=\ln \left(T_1\right)-\ln \left(T_2\right)\\ \Rightarrow \ln \left(\frac{2V}{V}\right)=\ln \left(\frac{T_1}{T_2}\right)\\ \Rightarrow T_2=\frac{T_1}{2}\end{array}$