Q24 / 150 · MHT CET 2026 (19 April Shift 1) · Chemistry
An ideal gas expands from initial volume $5\ \text{dm}^3$ to $15\ \text{dm}^3$ against a constant external pressure of $2$ atm. What is the work done by the gas ?
- A$-1202.6$ J
- Correct answerB$-2026.4$ J
- C$-1101.32$ J
- D$-1013.2$ J
Solution
The work done by a gas during expansion against a constant external pressure is given by the formula:
$W = -P_{\text{ext}} \Delta V$
Given:
$P_{\text{ext}} = 2\ \text{atm}$
$V_i = 5\ \text{dm}^3 = 5\ \text{L}$
$V_f = 15\ \text{dm}^3 = 15\ \text{L}$
Change in volume:
$\Delta V = V_f - V_i = 15 - 5 = 10\ \text{L}$
Substituting the values into the formula:
$W = -2\ \text{atm} \times 10\ \text{L} = -20\ \text{L}\ \text{atm}$
To convert the work done into Joules, use the conversion factor $1\ \text{L}\ \text{atm} = 101.32\ \text{J}$:
$W = -20 \times 101.32\ \text{J} = -2026.4\ \text{J}$
Answer: $-2026.4$ J
$W = -P_{\text{ext}} \Delta V$
Given:
$P_{\text{ext}} = 2\ \text{atm}$
$V_i = 5\ \text{dm}^3 = 5\ \text{L}$
$V_f = 15\ \text{dm}^3 = 15\ \text{L}$
Change in volume:
$\Delta V = V_f - V_i = 15 - 5 = 10\ \text{L}$
Substituting the values into the formula:
$W = -2\ \text{atm} \times 10\ \text{L} = -20\ \text{L}\ \text{atm}$
To convert the work done into Joules, use the conversion factor $1\ \text{L}\ \text{atm} = 101.32\ \text{J}$:
$W = -20 \times 101.32\ \text{J} = -2026.4\ \text{J}$
Answer: $-2026.4$ J
Hence, the correct answer is option: 'B'.