Question

One mole of an ideal gas is kept enclosed under a light piston ( area = ${10}^{-2}{m}^2$) connected by a compressed spring (spring constant 100 $N/m$). The volume of a gas is 0.83$m^3$ and its temperature is 100 $K$. The gas is heated so that it compresses the spring further by 0.1 $m$. The work done by the gas in the process is : (Take $R$ = 8.3 $J/(K$$-mol)$ and suppose there is no atmosphere).

• A. 3 $J$
• B. 6 $J$
• C. 9 $J$
• D. 1.5 $J$

Answer 1

The correct option is D 1.5 $J$

Before heating let the pressure of gas be $P$.
As the piston is in equilibrium ,
$PA=k{x}_1$
⇒ $x_1=\frac{PA}{K}=\left(\frac{nRT}{V}\right)\frac{A}{K}=\frac{1\times 8.3\times 100\times {10}^{-2}}{0.83\times 100}=0.1m$
Since, during heating process,
the spring is compressed further
by $0.1m$.
$\therefore x_2=0.2m$
Work done by gas will transfered in changing elastic potential energy of spring. Therefore work done by gas is
$\frac{1}{2}\times k({\left(x_2\right)}^2-{\left(x_1\right)}^2)=\frac{1}{2}\times 100({(0.2)}^2-{(0.1)}^2)=1.5J$