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Thermodynamics question

2024 · 27 Jan · Shift 1 · Q24
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Thermodynamics question

2024 · 27 Jan · Shift 1 · Q24

JEE MainChemistryThermodynamicsNumerical+4 / −1
If three moles of an ideal gas at 300 K300 \mathrm{~K}300 K expand isothermally from 30 dm330 \mathrm{~dm}^330 dm3 to 45 dm345 \mathrm{~dm}^345 dm3 against a constant opposing pressure of 80 kPa80 \mathrm{~kPa}80 kPa, then the amount of heat transferred is ‾\underline{\hspace{2cm}}​ J.
Numerical answer
View written solutionFree

Correct answer: 1200

  1. Given data
  • Number of moles: n=3n = 3n=3
  • Temperature: T=300 KT = 300\,\mathrm{K}T=300K
  • Initial volume: Vi=30 dm3V_i = 30\,\mathrm{dm^3}Vi​=30dm3
  • Final volume: Vf=45 dm3V_f = 45\,\mathrm{dm^3}Vf​=45dm3
  • Constant opposing pressure: Pext=80 kPaP_{\text{ext}} = 80\,\mathrm{kPa}Pext​=80kPa
  1. Use first law of thermodynamics

ΔU=q+w\Delta U = q + wΔU=q+w

For an ideal gas undergoing isothermal process, internal energy depends only on temperature. Since temperature is constant,

ΔU=0\Delta U = 0ΔU=0

Therefore,

q=−wq = -wq=−w

  1. Calculate work done

For expansion against constant external pressure,

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

Now,

ΔV=Vf−Vi=45−30=15 dm3\Delta V = V_f - V_i = 45 - 30 = 15\,\mathrm{dm^3}ΔV=Vf​−Vi​=45−30=15dm3

Since

1 kPa dm3=1 J1\,\mathrm{kPa\,dm^3} = 1\,\mathrm{J}1kPadm3=1J

we get

w=−(80)(15)=−1200 Jw = -(80)(15) = -1200\,\mathrm{J}w=−(80)(15)=−1200J

  1. Find heat transferred

q=−w=−(−1200)=1200 Jq = -w = -(-1200) = 1200\,\mathrm{J}q=−w=−(−1200)=1200J

  1. Final answer

The amount of heat transferred is

1200 J\boxed{1200\,\mathrm{J}}1200J​

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