SATHEE: Thermodynamics Question 51

Thermodynamics Question 51

Question: During an adiabatic expansion of 2 moles of a gas, the change in internal energy was found -50J. The work done during the process is

[Pb. PET 1996]

Options:

A) $350 \times 4^{0.4} K$

B) $300 \times 4^{0.4}$ K

C) $150 \times 4^{0.4}$ K

D) None of these

Show Answer

Answer:

Correct Answer: B

Solution:

For adiabatic change $T V^{γ - 1}$ = constant

Therefore $\frac{T_{2}}{T_{1}} = {(\frac{V_{1}}{V_{2}})}^{γ - 1}$

Therefore $T_{2} = {(\frac{V_{1}}{V_{2}})}^{γ - 1} \times T_{1}$

Therefore $T_{2} = {(\frac{V}{V / 4})}^{1.4 - 1} \times 300$

$= 300 \times {(4)}^{0.4} K$

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Thermodynamics Question 51

Question: During an adiabatic expansion of 2 moles of a gas, the change in internal energy was found -50J. The work done during the process is

Options:

Answer:

Solution:

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