Heat and Thermodynamics

To analyze the increase in internal energy when heat is absorbed by a monoatomic gas:First, we know that the heat absorbed, $Q$, is given as 40 J .For a monoatomic ideal gas, the specific heat capacities are:$\;C_p=\;\frac{5R}{2}$$\;C_V=\;\frac{3R}{2}$The relationship for heat transfer in terms of $C_p$ is:$Q=n C_p \Delta T$Substituting in known values:$40=n\cdot\;\frac{5R}{2}\cdot\Delta T$From this, we obtain:$nR\Delta T=16\;\;(i)$Next, we determine the increase in internal energy, $\Delta U$, which is given by:$\;\Delta U=n C_V \Delta T$$\;\Delta U=n\cdot\;\frac{3}{2}R\Delta T$$\;\Delta U=\;\frac{3}{2}\cdot nR\Delta T$Using Equation (i), we substitute for $nR\Delta T$ :$\Delta U=\;\frac{3}{2}\times 16$This simplifies to:$\Delta U=24\ \mathrm{J}$Thus, the increase in the internal energy of the gas is 24 J .