Question

The pressure and density of a diatomic gas $\left(\gamma =\frac{7}{5}\right)$ change adiabatically from $(P_1,{\rho }_1)$ to $(P_2{\rho }_2).$ IF $\frac{{\rho }_2}{{\rho }_1}=32,$ then $\frac{P_2}{P_1}$ should be:

• A. 16
• B. 32
• C. 64
• D. 128

Answer 1

Answer: (D)

In an adiabatic process. $P{V}^{\gamma }=\text{constant}$ or $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$ or $\frac{P_2}{P_1}={\left(\frac{V_1}{V_2}\right)}^{\gamma }$ ?(i) Volume of gas $=\frac{\text{Mass}}{\text{Density}}$ ie., $V=\frac{M}{\rho }$ or $V\propto \frac{1}{\rho }$ $\therefore$ $\frac{V_1}{V_2}=\frac{{\rho }_2}{{\rho }_1}=32$ Thus, form Eq.(i), we have $\frac{P_2}{P_1}={(32)}^{\gamma }={(32)}^{7/5}=2^7=128$