Question

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

• A. $3\text{J}$
• B. $6\text{J}$
• C. $9\text{J}$
• D. $1.5\text{J}$

Answer 1

The correct option is D $1.5\text{J}$

Before heating, let the pressure of the gas be $P$.
When the piston is at equilibrium, we can say that
$PA=k{x}_1$ {where $x_1$ is initial compression in the spring}
$\Rightarrow x_1=\frac{PA}{k}$
Using, $PV=nRT$, we get
$x_1=\left(\frac{nRT}{V}\right)\frac{A}{k}$
Substituting the values from the data given in the question,
$\Rightarrow x_1=\frac{1\times 8.3\times 100\times {10}^{-2}}{0.83\times 100}=0.1\text{m}$
Since, during heating process, the spring is compressed further by $0.1\text{m}$
$\therefore$ we get, $x_2=0.2\text{m}$
Since there is no work done by the atmosphere,
Work done by the gas = Change in P.E of spring
i.e Work done by the gas
$W=\frac{1}{2}\times k\times (x_2^2-x_1^2)$
$\Rightarrow W=\frac{1}{2}\times 100\times \left(\right(0.2{)}^2-\left(0.1{)}^2\right)$
$=\frac{1}{2}\times 100\times \left(0.1\right)\times \left(0.3\right)=1.5\text{J}$
Thus, option (d) is the correct answer.