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Heat and Thermodynamics question

2006 · Shift 0 · Q107
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Heat and Thermodynamics question

2006 · Shift 0 · Q107

JEE MainPhysicsHeat and ThermodynamicsMCQ+4 / −1
The work of 146kJ146kJ146kJ is performed in order to compress one kilo mole of gas adiabatically and in this process the temperature of the gas increases by 7∘C.{7^ \circ }C.7∘C. The gas is (R=8.3J  mol−1 K−1)\left( {R = 8.3J\,\,mo{l^{ - 1}}\,{K^{ - 1}}} \right)(R=8.3Jmol−1K−1)
  1. A
    diatomic
  2. B
    triatomic
  3. C
    a mixture of monoatomic and diatomic
  4. D
    monoatomic
View written solutionFree

Correct answer: A

  1. Use the first law for an adiabatic process

For an adiabatic compression, Q=0Q=0Q=0 So, ΔU=Won gas\Delta U = W_{\text{on gas}}ΔU=Won gas​

Given:

  • Work done on gas =146 kJ=146000 J=146\,\text{kJ}=146000\,\text{J}=146kJ=146000J
  • Amount of gas =1=1=1 kilomole =1000=1000=1000 moles
  • Rise in temperature =7∘C=7 K=7^\circ C = 7\,K=7∘C=7K

Thus, ΔU=nCVΔT\Delta U = nC_V\Delta TΔU=nCV​ΔT so 146000=1000 CV (7)146000 = 1000\, C_V\, (7)146000=1000CV​(7)

  1. Calculate molar heat capacity at constant volume

CV=1460001000×7C_V = \frac{146000}{1000\times 7}CV​=1000×7146000​ CV=1460007000C_V = \frac{146000}{7000}CV​=7000146000​ CV≈20.86 J mol−1K−1C_V \approx 20.86\,\text{J mol}^{-1}\text{K}^{-1}CV​≈20.86J mol−1K−1

  1. Compare with standard values

Using R=8.3 J mol−1K−1R=8.3\,\text{J mol}^{-1}\text{K}^{-1}R=8.3J mol−1K−1:

  • Monoatomic gas: CV=32R=12.45 J mol−1K−1C_V = \frac{3}{2}R = 12.45\,\text{J mol}^{-1}\text{K}^{-1}CV​=23​R=12.45J mol−1K−1

  • Diatomic gas: CV=52R=20.75 J mol−1K−1C_V = \frac{5}{2}R = 20.75\,\text{J mol}^{-1}\text{K}^{-1}CV​=25​R=20.75J mol−1K−1

  • Triatomic gas (non-linear, approximately): CV=3R=24.9 J mol−1K−1C_V = 3R = 24.9\,\text{J mol}^{-1}\text{K}^{-1}CV​=3R=24.9J mol−1K−1

The calculated value CV≈20.86 J mol−1K−1C_V \approx 20.86\,\text{J mol}^{-1}\text{K}^{-1}CV​≈20.86J mol−1K−1 is closest to 52R=20.75 J mol−1K−1\frac{5}{2}R = 20.75\,\text{J mol}^{-1}\text{K}^{-1}25​R=20.75J mol−1K−1

  1. Conclusion

Hence, the gas is diatomic.

Therefore, the correct option is: A\boxed{\text{A}}A​

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