Question

500 J of energy is transferred as heat to 0.5 mol of Argon gas at 298 K and 1.00 atm. The final temperature and the change in internal energy respectively are: Given: $R = 8.3\ \text{J K}^{-1}\text{mol}^{-1}$

• A. 378 K and 500 J
• B. 368 K and 500 J
• C. 348 K and 300 J
• D. 378 K and 300 J

Hint:

At constant pressure, part of the supplied heat is used to do work. Hence $\Delta U < Q$.

Answer 1

The Correct Option is C

Concept:

Argon is a monoatomic ideal gas.
At constant pressure, heat supplied is given by: \[ Q = n C_p \Delta T \]
Change in internal energy is: \[ \Delta U = n C_v \Delta T \]
For monoatomic gas: \[ C_v = \frac{3}{2}R, \quad C_p = \frac{5}{2}R \]
Step 1: Given data: \[ Q = 500\ \text{J}, \quad n = 0.5\ \text{mol}, \quad T_i = 298\ \text{K} \]
Step 2: Calculate $C_p$: \[ C_p = \frac{5}{2}R = \frac{5}{2} \times 8.3 = 20.75\ \text{J mol}^{-1}\text{K}^{-1} \]
Step 3: Temperature rise: \[ \Delta T = \frac{Q}{n C_p} = \frac{500}{0.5 \times 20.75} \] \[ \Delta T \approx 48\ \text{K} \]
Step 4: Final temperature: \[ T_f = T_i + \Delta T = 298 + 48 = 346 \approx 348\ \text{K} \]
Step 5: Calculate $C_v$: \[ C_v = \frac{3}{2}R = 12.45\ \text{J mol}^{-1}\text{K}^{-1} \]
Step 6: Change in internal energy: \[ \Delta U = n C_v \Delta T = 0.5 \times 12.45 \times 48 \] \[ \Delta U \approx 300\ \text{J} \]