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Thermodynamics question

2019 · 12 Apr · Shift 1 · Q12
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Thermodynamics question

2019 · 12 Apr · Shift 1 · Q12

JEE MainChemistryThermodynamicsMCQ+4 / −1
An ideal gas is allowed to expand form 1 L to 10 L against a constant external pressure of I bar. The work done in kJ is :
  1. A
    +10.0
  2. B
    –0.9
  3. C
    – 2.0
  4. D
    – 9.0
View written solutionFree

Correct answer: B

  1. Use the formula for work at constant external pressure

For expansion against a constant external pressure,

w=−Pext ΔVw = -P_{\text{ext}}\,\Delta Vw=−Pext​ΔV

where

ΔV=V2−V1\Delta V = V_2 - V_1ΔV=V2​−V1​
  1. Substitute the given values

Initial volume: V1=1 LV_1 = 1\,\text{L}V1​=1L Final volume: V2=10 LV_2 = 10\,\text{L}V2​=10L So,

ΔV=10−1=9 L\Delta V = 10 - 1 = 9\,\text{L}ΔV=10−1=9L

External pressure:

Pext=1 barP_{\text{ext}} = 1\,\text{bar}Pext​=1bar

Thus,

w=−(1 bar)(9 L)=−9 L⋅barw = -(1\,\text{bar})(9\,\text{L}) = -9\,\text{L·bar}w=−(1bar)(9L)=−9L⋅bar
  1. Convert L⋅bar\text{L·bar}L⋅bar into kJ

We use:

1 L⋅bar=100 J=0.1 kJ1\,\text{L·bar} = 100\,\text{J} = 0.1\,\text{kJ}1L⋅bar=100J=0.1kJ

Therefore,

−9 L⋅bar=−9×0.1=−0.9 kJ-9\,\text{L·bar} = -9 \times 0.1 = -0.9\,\text{kJ}−9L⋅bar=−9×0.1=−0.9kJ
  1. Match with the options

The work done is

−0.9 kJ\boxed{-0.9\,\text{kJ}}−0.9kJ​

So the correct option is B.

  1. Comparison with stored answer

Stored correct answer: B

Our derived answer also gives B, so they agree.

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