Thermodynamics Part 3

Given, the number of diatomic moles, n = 3 mol The increase in temperature of the diatomic mole, ΔT = 40ºC Now, the degree of freedom, f = linear + rotational + no oscillation f = 3 + 2 + 0 f = 5 Change in internal energy, $\Delta U = n C_V \Delta T$ where, $C_V = \frac{f}{2} R = \frac{5}{2} R$ Substituting the value in Eq. (i), we get $\Delta U = \frac{5R}{2} n \Delta T$ Now, work done by the gas for isobaric process, $W = p \Delta V = n R \Delta T$ The ratio of the change in internal energy to the work done by thegas, $\frac{\Delta U}{W} = \frac{\frac{5}{2} n R \Delta T}{n R \Delta T}$ $\frac{\Delta U}{W} = \frac{5}{2}$ Multiply and divide the above equation with 5, we get $\frac{\Delta U}{W} = \frac{5 \times 5}{2 \times 5} = \frac{25}{10}$ Comparing with given equation, $\frac{\Delta U}{W} = \frac{x}{10}$ The value of the x = 25.