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Thermodynamics question

2019 · 10 Jan · Shift 2 · Q14
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Thermodynamics question

2019 · 10 Jan · Shift 2 · Q14

JEE MainChemistryThermodynamicsMCQ+4 / −1
An ideal gas undergoes isothermal compression from 5m3 to 1 m3 against a constant external pressure of 4 Nm–2. Heat released in this process is used to increase the temperature of 1 mole of Al. If molar heat capacity of Al is 24 J mol–1 K–1, the temperature of Al increases by :
  1. A
    23K{2 \over 3}K32​K
  2. B
    32K{3 \over 2}K23​K
  3. C
    1 K
  4. D
    2 K
View written solutionFree

Correct answer: A

  1. Given data
  • Isothermal compression of an ideal gas
  • Initial volume: Vi=5 m3V_i = 5\,\text{m}^3Vi​=5m3
  • Final volume: Vf=1 m3V_f = 1\,\text{m}^3Vf​=1m3
  • Constant external pressure: Pext=4 N m−2P_{\text{ext}} = 4\,\text{N m}^{-2}Pext​=4N m−2
  • Heat released is used to heat 111 mole of Al
  • Molar heat capacity of Al: C=24 J mol−1K−1C = 24\,\text{J mol}^{-1}\text{K}^{-1}C=24J mol−1K−1
  1. Work done during compression

For compression against constant external pressure,

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

Substitute values:

w=−4(1−5)=−4(−4)=16 Jw = -4(1 - 5) = -4(-4) = 16\,\text{J}w=−4(1−5)=−4(−4)=16J

So, work done on the gas is 16 J16\,\text{J}16J.

  1. Use first law of thermodynamics

For an ideal gas undergoing isothermal process,

ΔU=0\Delta U = 0ΔU=0

Hence,

ΔU=q+w=0\Delta U = q + w = 0ΔU=q+w=0

q=−w=−16 Jq = -w = -16\,\text{J}q=−w=−16J

So the gas releases heat of magnitude

∣q∣=16 J|q| = 16\,\text{J}∣q∣=16J

  1. Heat absorbed by aluminium

This released heat is used to raise the temperature of 111 mole of Al.

Q=nCΔTQ = nC\Delta TQ=nCΔT

Here,

16=1×24×ΔT16 = 1 \times 24 \times \Delta T16=1×24×ΔT

ΔT=1624=23 K\Delta T = \frac{16}{24} = \frac{2}{3}\,\text{K}ΔT=2416​=32​K

  1. Final answer

The temperature increase of aluminium is

23 K\boxed{\frac{2}{3}\,\text{K}}32​K​

So, Option A is correct.

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