A total of $48 \mathrm{~J}$ heat is given to one mole of helium kept in a cylinder. The temperature of helium increases by $2^{\circ} \mathrm{C}$. The work done by the gas is: Given, $\mathrm{R}=8.3 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$.

<p>To solve this problem, we can use the first law of thermodynamics, which states:</p> <p>$\Delta Q = \Delta U + W$</p> <p>where:</p> <ul> <li>$\Delta Q$ is the heat transferred to the system, in this case, $48 \mathrm{~J}$. </li> <li>$\Delta U$ is the change in internal energy of the system. </li> <li>$W$ is the work done by the system.</li> </ul> <p>For one mole of an ideal gas, the change in internal energy can be calculated using the equation:</p> <p>$\Delta U = nC_{v}\Delta T$</p> <p>where:</p> <ul> <li>$n$ is the number of moles, which is $1$ in this case. </li> <li>$C_{v}$ is the molar specific heat capacity at constant volume. </li> <li>$\Delta T$ is the change in temperature, in Kelvins.</li> </ul> <p>Given that we are dealing with helium, a monatomic ideal gas, we know that:</p> <p>$C_{v} = \frac{3}{2}R$</p> <p>Substituting the given values, we get:</p> <p>$$\Delta U = (1) \frac{3}{2}(8.3 \mathrm{~J} \mathrm{~mol}^{-1} \mathrm{~K}^{-1})(2^{\circ} \mathrm{C})$$</p> <p>Since a change in temperature of $1^{\circ} \mathrm{C}$ is equivalent to a change in temperature of $1 \mathrm{~K}$, we can directly substitute $2^{\circ} \mathrm{C}$ as a $2 \mathrm{~K}$ change without needing to convert Celsius to Kelvin as both scales have the same magnitude of degree.</p> <p>Therefore,</p> <p>$\Delta U = 1 \times \frac{3}{2} \times 8.3 \times 2 = 24.9 \mathrm{~J}$</p> <p>Now, substituting $\Delta U$ and $\Delta Q$ into the first law of thermodynamics:</p> <p>$48 = 24.9 + W$</p> <p>Solving for $W$ we find:</p> <p>$W = 48 - 24.9 = 23.1 \mathrm{~J}$</p> <p>Thus, the work done by the gas is $23.1 \mathrm{~J}$, which corresponds to Option A.</p>