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Thermodynamics question

2024 · 4 Apr · Shift 2 · Q27
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Thermodynamics question

2024 · 4 Apr · Shift 2 · Q27

JEE MainChemistryThermodynamicsNumerical+4 / −1
Three moles of an ideal gas are compressed isothermally from 60 L60 \mathrm{~L}60 L to 20 L20 \mathrm{~L}20 L using constant pressure of 5 atm5 \mathrm{~atm}5 atm. Heat exchange Q\mathrm{Q}Q for the compression is - ‾\underline{\hspace{2cm}}​ Lit. atm.
Numerical answer
View written solutionFree

Correct answer: 200

  1. Given data
  • Number of moles: n=3n=3n=3
  • Initial volume: Vi=60 LV_i=60\,\text{L}Vi​=60L
  • Final volume: Vf=20 LV_f=20\,\text{L}Vf​=20L
  • Process: isothermal compression
  • External constant pressure used for compression: Pext=5 atmP_{\text{ext}}=5\,\text{atm}Pext​=5atm

We need the heat exchanged QQQ in units of L-atm.

  1. Use first law of thermodynamics

For any process,

ΔU=Q+W\Delta U = Q + WΔU=Q+W

where WWW is work done on the system.

For an ideal gas undergoing an isothermal process,

ΔU=0\Delta U = 0ΔU=0

because internal energy depends only on temperature.

Hence,

0=Q+W0 = Q + W0=Q+W

so,

Q=−WQ = -WQ=−W
  1. Calculate work done on the gas

Since compression occurs against constant external pressure,

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

Substitute the values:

W=5×(60−20)W = 5\times(60-20)W=5×(60−20) W=5×40=200 L-atmW = 5\times 40 = 200\,\text{L-atm}W=5×40=200L-atm
  1. Find heat exchanged

Since

Q=−WQ=-WQ=−W

we get

Q=−200 L-atmQ=-200\,\text{L-atm}Q=−200L-atm

The question asks: Heat exchange QQQ for the compression is −‾-\underline{\hspace{2cm}}−​ L-atm.

So the blank should contain:

200200200
  1. Final answer
Q=−200 L-atmQ=-200\,\text{L-atm}Q=−200L-atm

Therefore, the required integer is:

200\boxed{200}200​
  1. Comparison with stored correct answer

Stored correct answer = 200200200

This matches our derived result for the blank.

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