Question

$n$ moles of a gas are filled in a container at temperature $T$ is in equilibrium initially. if the gas is slowly and isothermally compressed to half its initial volume the work done by the atmosphere on the gas is

• A. $\frac{nRT}{2}$
• B. $-\frac{nRT}{2}$
• C. $nRT(\ell n2-\frac{1}{2})$
• D. $-nRTln2$

Answer 1

The correct option is C $nRT(\ell n2-\frac{1}{2})$

Work done is given by:
$w=\int PdV$
$PV=nRT$
$P=\frac{1}{V}nRT$
i.e., $w=\int \frac{1}{V}nRTdV=nRT\ln \left(\frac{V_2}{V_1}\right)$
Given that: $V_2=\frac{V_1}{2}$
$w=-nRT\ln 2$
The work done by atmospheric pressure is equal to $-w$
$w_{atm}=nRT\ln 2$