Question

A sample consisting of 2.00 mol of perfect gas molecules at 250 K is compressed reversibly and adiabatically until its temperature reaches 300 K. Given that CV,m = 27.5 J K−1 mol−1 , calculate q, w, ΔU, ΔH, and ΔS. (Hint, assume heat capacity CV,m is a constant during the compression.)

Answer 1

The calculation involves determining the change in internal energy, enthalpy, and work for a reversible adiabatic compression of a perfect gas, with changes in entropy being zero.

Step-by-step explanation:

The student's question regards the calculation of thermodynamic properties such as heat (q), work (w), change in internal energy (ΔU), change in enthalpy (ΔH), and change in entropy (ΔS) during a reversible and adiabatic compression of a perfect gas. For an adiabatic process, q is zero because no heat is exchanged with the surroundings. The work done on the system, w, is equal to the change in internal energy, ΔU, which can be calculated using ΔU = nCV,mΔT. The change in enthalpy, ΔH, can be found using ΔH = nCp,mΔT, where Cp,m = CV,m + R (the gas constant R), and since this is a reversible adiabatic process, the change in entropy, ΔS, is zero.