Q122 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Chemistry
Calculate the work done when an ideal gas expands from $1 \times 10^{-3}$ to $1 \times 10^{-2}\ \text{m}^3$ at 300 K against constant pressure of $1 \times 10^5\ \text{Nm}^{-2}$. ?
- A900kJ
- B$-900$kJ
- C270kJ
- Correct answerD$-900$J
Solution
Work done during expansion against a constant external pressure is given by $W = -P_{\text{ext}} \Delta V$.
Given $P_{\text{ext}} = 1 \times 10^5\ \text{Nm}^{-2}$.
Initial volume $V_1 = 1 \times 10^{-3}\ \text{m}^3$.
Final volume $V_2 = 1 \times 10^{-2}\ \text{m}^3$.
Change in volume $\Delta V = V_2 - V_1 = 1 \times 10^{-2} - 1 \times 10^{-3} = 9 \times 10^{-3}\ \text{m}^3$.
Substituting the values:
$W = -1 \times 10^5 \times 9 \times 10^{-3} = -900\ \text{J}$.
Answer: $-900$J
Given $P_{\text{ext}} = 1 \times 10^5\ \text{Nm}^{-2}$.
Initial volume $V_1 = 1 \times 10^{-3}\ \text{m}^3$.
Final volume $V_2 = 1 \times 10^{-2}\ \text{m}^3$.
Change in volume $\Delta V = V_2 - V_1 = 1 \times 10^{-2} - 1 \times 10^{-3} = 9 \times 10^{-3}\ \text{m}^3$.
Substituting the values:
$W = -1 \times 10^5 \times 9 \times 10^{-3} = -900\ \text{J}$.
Answer: $-900$J
Hence, the correct answer is option: 'D'.