The Molar Specific Heats of an Ideal Gas — Question 8

Question 8

Equal heat inputs do not imply equal final ideal-gas temperatures

Use $pV=nRT$, $R=8.314$ J/(mol K), and the sign convention $\Delta U=Q-W$, with $W$ the work done by the gas. Only pressure–volume work occurs. Two identical .200 mol samples begin at 300 K with $C_V=\tfrac32R$. Each receives 249.42 J reversibly. A is held at fixed volume; B is held at fixed pressure.

1. Find both final temperatures and the fraction of B heat stored internally.

2. If the two initial volumes are equal, compare the final pressures and volumes relative to their common initial values.

Question 8 - Solution

Equal heat inputs do not imply equal final ideal-gas temperatures

(a) Method and calculation

$$\Delta T_A=\frac{249.42}{(.200)(1.50)(8.314)}=100\,\mathrm K\ \Rightarrow\ T_A=400\,\mathrm K.$$$$\Delta T_B=\frac{249.42}{(.200)(2.50)(8.314)}=60.0\,\mathrm K\ \Rightarrow\ T_B=360\,\mathrm K.$$$$\Delta U_B/Q_B=C_V/C_p=3/5.$$

(b) Interpretation and further calculation

$$p_A/p_i=T_A/T_i=4/3,\quad V_A/V_i=1.$$$$p_B/p_i=1,\quad V_B/V_i=T_B/T_i=6/5.$$B spends 40% of its heat on expansion, reducing its temperature rise relative to A.

Physical check

The same heat input creates different final states because the mechanical constraints differ.