Question

The heat supplied to an ideal diatomic gas so that it expands from a volume of $18 \times 10^{-3} m^3$ to $37 \times 10^{-3} m^3$ at a constant pressure of 2 atmospheres is (Atmospheric pressure = $10^5$ Pa)

• A. 9500 J
• B. 3800 J
• C. 5700 J
• D. 13300 J

Hint:

Isobaric process shortcuts: - Monatomic: $Q = \frac{5}{2} P \Delta V$ - Diatomic: $Q = \frac{7}{2} P \Delta V$

Answer 1

The Correct Option is D

Step 1: Identify the Process:
The expansion is at constant pressure, so it is an Isobaric Process. Heat supplied at constant pressure is given by: \[ Q_p = n C_p \Delta T \] Using ideal gas equation $PV = nRT$, for isobaric process $P \Delta V = nR \Delta T \Rightarrow n \Delta T = \frac{P \Delta V}{R}$. Substitute into heat equation: \[ Q_p = C_p \left( \frac{P \Delta V}{R} \right) = \frac{C_p}{R} P \Delta V \]
Step 2: Specific Heat for Diatomic Gas:
For a diatomic gas, molar heat capacity at constant pressure is: \[ C_p = \frac{7}{2} R \] So, \[ Q_p = \frac{\frac{7}{2} R}{R} P \Delta V = \frac{7}{2} P \Delta V \]
Step 3: Substitution and Calculation:
Pressure $P = 2$ atm $= 2 \times 10^5$ Pa. Change in Volume $\Delta V = V_2 - V_1 = (37 - 18) \times 10^{-3} m^3 = 19 \times 10^{-3} m^3$. \[ Q_p = \frac{7}{2} \times (2 \times 10^5) \times (19 \times 10^{-3}) \] \[ Q_p = 7 \times 10^5 \times 19 \times 10^{-3} \] \[ Q_p = 7 \times 19 \times 10^2 \] \[ Q_p = 133 \times 100 = 13300 \, \text{J} \]
Step 4: Final Answer:
The heat supplied is 13300 J.