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 $