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Thermodynamics question

2004 · Shift 0 · Q1
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Thermodynamics question

2004 · Shift 0 · Q1

JEE MainChemistryThermodynamicsMCQ+4 / −1
An ideal gas expands in volume from 1 ×\times× 10-3 m3 to 1 ×\times× 10-2 m3 at 300 K against a constant pressure of 1 ×\times× 105 Nm-2. The work done is :
  1. A
    -900 J
  2. B
    900 kJ
  3. C
    270 kJ
  4. D
    -900 kJ
View written solutionFree

Correct answer: A

  1. Given data
  • Initial volume: Vi=1×10−3 m3V_i = 1\times 10^{-3}\,\text{m}^3Vi​=1×10−3m3
  • Final volume: Vf=1×10−2 m3V_f = 1\times 10^{-2}\,\text{m}^3Vf​=1×10−2m3
  • External constant pressure: Pext=1×105 N m−2P_{\text{ext}} = 1\times 10^5\,\text{N m}^{-2}Pext​=1×105N m−2
  1. Formula for work done in expansion against constant external pressure

For expansion,

w=−Pext(Vf−Vi)w = -P_{\text{ext}}(V_f - V_i)w=−Pext​(Vf​−Vi​)

The negative sign is used because, by chemistry sign convention, work done by the system during expansion is negative.

  1. Calculate change in volume
ΔV=Vf−Vi=1×10−2−1×10−3\Delta V = V_f - V_i = 1\times 10^{-2} - 1\times 10^{-3}ΔV=Vf​−Vi​=1×10−2−1×10−3 ΔV=0.01−0.001=0.009 m3=9×10−3 m3\Delta V = 0.01 - 0.001 = 0.009\,\text{m}^3 = 9\times 10^{-3}\,\text{m}^3ΔV=0.01−0.001=0.009m3=9×10−3m3
  1. Calculate work
w=−(1×105)(9×10−3)w = -(1\times 10^5)(9\times 10^{-3})w=−(1×105)(9×10−3) w=−9×102 Jw = -9\times 10^2\,\text{J}w=−9×102J w=−900 Jw = -900\,\text{J}w=−900J
  1. Check options
  • A: −900 J-900\,\text{J}−900J ✅
  • B: 900 kJ900\,\text{kJ}900kJ ❌
  • C: 270 kJ270\,\text{kJ}270kJ ❌
  • D: −900 kJ-900\,\text{kJ}−900kJ ❌

Therefore, the correct option is A.

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