The Molar Specific Heats of an Ideal Gas — Question 9

Question 9

A two-step heating experiment checks energy conservation across paths

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. One mole with $C_V=\tfrac52R$ starts at $(p_0,V_0,T_0)$, where $T_0=300$ K. It first doubles pressure at fixed volume, then doubles volume at the new constant pressure.

1. Find the temperature, heat and work for each step and the total internal-energy change.

2. Compare with a hypothetical single fixed-volume temperature rise from $T_0$ to the same final temperature. Explain what is identical and what is different, including whether the final states match.

Question 9 - Solution

A two-step heating experiment checks energy conservation across paths

(a) Method and calculation

$$T_1=2T_0=600\,\mathrm K,\quad T_2=4T_0=1200\,\mathrm K.$$$$W_1=0,\quad Q_1=\tfrac52R(300)=6235.50\,\mathrm J.$$$$W_2=R(600)=4988.40\,\mathrm J,\quad Q_2=\tfrac72R(600)=17459.40\,\mathrm J.$$

(b) Interpretation and further calculation

$$Q_{\mathrm{total}}=23694.90\,\mathrm J,\quad\Delta U=Q-W=18706.50\,\mathrm J=\tfrac52R(900).$$$$Q_{\mathrm{rigid}}=18706.50\,\mathrm J,\quad W_{\mathrm{rigid}}=0.$$Internal energies match at the common final temperature. Final states differ: the two-step endpoint is $(2p_0,2V_0)$, while the rigid endpoint is $(4p_0,V_0)$.

Physical check

For an ideal gas, equal temperatures imply equal internal energy for equal amounts, even at different volumes.