Question:

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 ?

Show Hint

For a constant external pressure, w = minus P times the change in volume; convert L atm to joules.
Updated On: Oct 1, 2026
  • \(-1202.6\) J
  • \(-2026.4\) J
  • \(-1101.32\) J
  • \(-1013.2\) J
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is B

Solution and Explanation

Step 1: Understanding the Concept:
Expansion against a constant external pressure gives \(w = -P_{ext}\Delta V\). The negative sign means the system loses energy as work.

Step 2: Find the Volume Change:
\(\Delta V = 15 - 5 = 10\ \text{dm}^3 = 10\ \text{L}\).

Step 3: Calculate the Work:
\[ w = -2\ \text{atm} \times 10\ \text{L} = -20\ \text{L atm} \]
Use \(1\ \text{L atm} = 101.32\) J.
\[ w = -20 \times 101.32 = -2026.4\ \text{J} \]

Step 4: Check the Other Options:
\(-1013.2\) J corresponds to only 10 L atm, as if the pressure were 1 atm. \(-1202.6\) J and \(-1101.32\) J do not match any sensible product of P, \(\Delta V\) and 101.32. So (B) is correct.

Final Answer:
The work done is \(-2026.4\) J, option (B). \[ \boxed{\text{(B) } -2026.4\ \text{J}} \]
Was this answer helpful?
0
0