Question

Under isothermal conditions a gas expands from 0.2 dm³ to 0.8 dm³ against a constant pressure of 2 bar at 300 K. Find the work done by the gas.(1 dm³ bar = 100 J)

• A. 160 J
• B. –120 J
• C. –40 J
• D. 20 J

Answer 1

The Correct Option is B

The work done by a gas during expansion or compression: 
Work = - PΔV 
Here,
ΔV = final volume - initial volume 
ΔV = 0.8 dm³ - 0.2 dm³ 
ΔV = 0.6 dm³ 
The pressure (P) is given as 2 bar, which we can convert to the corresponding units of dm³ bar: 
1 dm³ bar = 100 J 
So, 2 bar = 200 J. 
Substituting the values into the formula, we get: 
Work = - PΔV 
Work = -(200 J) x (0.6 dm³) 
Work = -120 J 
Therefore, the work done by the gas is -120 J, which matches option (B) in the given choices.

Answer 2

Given:
Initial volume, \(V_i = 0.2\  dm^3\) (converted to \(m^3\)),
Final volume, \(V_f = 0.8\) dm³ (converted to m³),
External pressure, \(P_{\text{ext}} = 2\) bar,
Conversion factor: \(1\ dm^3\ bar = 100\) J.

Convert volumes from dm³ to m³:
\(V_i = 0.2 \, \text{dm}^3 = 0.2 \times 10^{-3} \, \text{m}^3 = 0.0002 \, \text{m}^3\)
\(V_f = 0.8 \, \text{dm}^3 = 0.8 \times 10^{-3} \, \text{m}^3 = 0.0008 \, \text{m}^3\)

Calculate the volume change:
\(\Delta V = V_f - V_i = 0.0008 \, \text{m}^3 - 0.0002 \, \text{m}^3 = 0.0006 \, \text{m}^3\)

Convert the external pressure from bar to Pa:
\(P_{\text{ext}} = 2 \times 10^5 \, \text{Pa}\) (since 1 bar = 10⁵ Pa)

Calculate the work done by the gas:
\(W = -P_{\text{ext}} \Delta V\)
\(W = -2 \times 10^5 \, \text{Pa} \times 0.0006 \, \text{m}^3\)
\(W = -120 \, \text{J}\)

So, the correct option is (B): -120J.