Question

Five moles of an ideal monatomic gas with an initial temperature of $150^\circ\text{C}$ expand and in the process absorb 1500 J of heat and does 2500 J of work. The final temperature of the gas in $^\circ\text{C}$ is (Ideal gas constant $R=8.314\,\text{J K}^{-1}\text{mol}^{-1}$)

• A. 134
• B. 126
• C. 144
• D. 166
• E. 174

Hint:

For a monatomic gas, $C_v = 1.5R$. If the work done by the gas is greater than the heat absorbed, the internal energy must drop, leading to a decrease in temperature.

Answer 1

The Correct Option is A

Concept:
Using the First Law of Thermodynamics: $\Delta U = Q - W$. For a monatomic ideal gas, the internal energy change is $\Delta U = n C_v \Delta T = n (\frac{3}{2}R) \Delta T$.

Step 1: Calculate the change in internal energy ($\Delta U$).
Given $Q = +1500$ J (absorbed) and $W = +2500$ J (done by the gas). \[ \Delta U = 1500 - 2500 = -1000 \text{ J} \]

Step 2: Calculate the change in temperature ($\Delta T$).
\[ -1000 = n \left(\frac{3}{2}R\right) \Delta T \] \[ -1000 = 5 \times \left(\frac{3}{2} \times 8.314\right) \Delta T \] \[ -1000 = 5 \times 12.471 \times \Delta T \implies -1000 = 62.355 \times \Delta T \] \[ \Delta T = \frac{-1000}{62.355} \approx -16.03 \text{ K (or } ^\circ\text{C)} \]

Step 3: Find the final temperature.
Initial temperature $T_i = 150^\circ$C. \[ T_f = T_i + \Delta T = 150 - 16.03 \approx 133.97^\circ\text{C} \] Rounding to the nearest whole number gives 134$^\circ$C.