Q176 / 180 · COMEDK 2025 (10 May Shift 2) · Physics
The heat required to increase the temperature of 4 moles of a mono-atomic ideal gas from $273^{\circ} \mathrm{C}$ to $473^{\circ} \mathrm{C}$ at constant volume is
- Correct answerA1200 R
- B600 R
- C1600 R
- D400 R
Solution
The heat required at constant volume is given by the formula $Q = n C_{v} \Delta T$.
For a mono-atomic ideal gas, the molar heat capacity at constant volume is $C_{v} = \dfrac{3}{2} R$.
Given values are $n = 4$ moles, $T_{1} = 273^{\circ} \text{C}$, and $T_{2} = 473^{\circ} \text{C}$.
The change in temperature is $\Delta T = T_{2} - T_{1} = 473 - 273 = 200 \text{ K}$.
Substituting these values into the formula:
$Q = 4 \times \left( \dfrac{3}{2} R \right) \times 200$
$Q = 4 \times 1.5 R \times 200$
$Q = 6 R \times 200 = 1200 R$.
Answer: 1200 R
For a mono-atomic ideal gas, the molar heat capacity at constant volume is $C_{v} = \dfrac{3}{2} R$.
Given values are $n = 4$ moles, $T_{1} = 273^{\circ} \text{C}$, and $T_{2} = 473^{\circ} \text{C}$.
The change in temperature is $\Delta T = T_{2} - T_{1} = 473 - 273 = 200 \text{ K}$.
Substituting these values into the formula:
$Q = 4 \times \left( \dfrac{3}{2} R \right) \times 200$
$Q = 4 \times 1.5 R \times 200$
$Q = 6 R \times 200 = 1200 R$.
Answer: 1200 R
Hence, the correct answer is option: 'A'.