Question

The pressure and density of a diatomic gas $ \left(\gamma=\dfrac{7}{5}\right) $ change adiabatically from $ \left(P_{1}, \rho_{1}\right) $ to $ \left(P_{2}, \rho_{2}\right) . $ If $ \dfrac{\rho_{2}}{\rho_{1}}=32, $ then $ \dfrac{P_{2}}{P_{1}} $ should be
(A) 126
(B) 128
(C) 146
(D) 124

Answer 1

Hint
We know that an adiabatic process occurs without transferring heat or mass between a thermodynamic system and its surroundings. Unlike an isothermal process, an adiabatic process transfers energy to the surroundings only as work. Heating in a called adiabatic condition occurs when the pressure of a gas is increased by the work done on it. An example of adiabatic heating is that of a heat engine piston that compresses a gas contained within a cylinder. Gas compression leads to a temperature rise. Adiabatic processes are not fast generally, if we have well insulated piston cylinder arrangement then we can obtain a very slow adiabatic process, if the pressure available in the cylinder causes a small net force to move the load.

Complete step by step answer
We know that,
For an adiabatic process,
 $ \mathrm{PV}^{\gamma}= $ constant
 $ \mathrm{P} \rho^{-\gamma}= $ constant
Therefore, we can equate,
 $ \dfrac{{{\text{P}}_{2}}}{{{\text{P}}_{1}}}={{\left( \dfrac{{{\rho }_{1}}}{{{\rho }_{2}}} \right)}^{-\gamma }} $
 $ ={{\left( \dfrac{1}{32} \right)}^{-7/5}} $
 $ ={{(32)}^{7/5}} $ $ \dfrac{\mathrm{P}_{2}}{\mathrm{P}_{1}}=\left(\dfrac{\rho_{1}}{\rho_{2}}\right)^{-\gamma}=\left(\dfrac{1}{32}\right)^{-7 / 5}=(32)^{7 / 5}=128 $
Therefore, the correct answer is Option (B).

Note
We know that a process without transfer of heat or matter to or from a system, so that Q = 0, is called adiabatic, and such a system is said to be adiabatically isolated. The assumption that a process is adiabatic is a frequently made simplifying assumption. An adiabatic process is one in which no heat is gained or lost by the system. The first law of thermodynamics with $Q=0$ shows that all the change in internal energy is in the form of work done. Entropy remains constant in an adiabatic process which is also reversible. Now, since the process is adiabatic, so the heat transfer is zero and so the entropy change is zero through heat transfer.