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Q. Two samples $A$ and $B$ of a gas, initially at the same pressure and temperature, are compressed from volume $V$ to $V/2$ ($A$ isothermally and $B$ adiabatically) the final pressure of $A$ is

Chhattisgarh PMTChhattisgarh PMT 2005

Solution:

For sample $A \rightarrow$ in isothermal process By Boyle's law
$p_{1} V_{1}=p_{2} V_{2}$
$\Rightarrow p_{2}=p\left(\frac{V_{1}}{V_{2}}\right)$
$p_{2}=p\left(\frac{V}{V / 2}\right)=2 p$
For sample $B \rightarrow$ in adiabatic process By Poisson's law $p V_{1}^{\gamma}=p_{2} V_{2}^{\gamma}$
$\therefore p_{2}=p(2)^{\gamma}$
$\because \gamma>1$
$\therefore 2 \gamma>2$
$\therefore 2 \gamma p>2 p$
or $\left(p_{2}\right)_{B}>\left(p_{2}\right)_{A}$
Final pressure of $A$ is less than pressure of $B$.