There are two samples $A$ and $B$ of a certain gas, which are initially at the same temperature and pressure…

There are two samples $A$ and $B$ of a certain gas, which are initially at the same temperature and pressure. Both are compressed from volume v to $\frac{\mathrm{v}}{2}$. Sample A is compressed isothermally while sample B is compressed adiabatically. The final pressure of $A$ is
  1. twice that of B.
  2. equal to that of $B$.
  3. more than that of $B$.
  4. less than that of B.

Solution

For an isothermal process. $P_1 V_1=P_2 V_2$
Given that $\mathrm{V}_2=\frac{\mathrm{V}_1}{2}$. It can be clearly understood that for $\mathrm{P}_2 \mathrm{~V}_2$ to remain constant, $\mathrm{P}_2=2 \mathrm{P}_1$. For an adiabatic process, $\mathrm{PV}^\gamma=$ constant. $\mathrm{P}_1 \mathrm{~V}_1{ }^\gamma=\mathrm{P}_2 \mathrm{~V}_2{ }^\gamma$ $P_2=P_1\left(\frac{V_1}{V_2}\right)^\gamma=2 P_1^\gamma$ $\gamma$ is always greater than 1. From equations (i) and (ii), it can be seen that the pressure change in an adiabatic process would be greater than that in an isothermal process for the same change in volume. :

Asked in: MHT CET 2024 (03 May Shift 1)

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