One mole He gas at 300K, 10 atm is heated through a process in which its pressure is directly proportional to square of its absolute temperature till its temperature becomes 600 K . The change in entr

Question

One mole He gas at 300K, 10 atm is heated through a process in which its pressure is directly proportional to square of its absolute temperature till its temperature becomes 600 K . The change in entropy of He is :-

TestHub · Accepted Answer

Question Explanation: Find the change in entropy of He.

Concept: Calculation of entropy of an ideal gas.

Solution:

$P \frac{P _{1}}{P _{2}} Δ S \propto T^{2} = (\frac{T _{1}}{T _{2}})^{2} = \frac{1}{4} = n C_{p} ln (\frac{T _{2}}{T _{1}}) + n R ln (\frac{P _{1}}{P _{2}}) = 1 \times \frac{5}{2} R ln 2 + 1 \times R ln (\frac{1}{4}) = \frac{R ln 2}{2}$

Final Answer: Option (B)

Chemistry - Thermodynamics - II Question with Solution | TestHub

Options:

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JEE Advanced 2027 (Class 12+)

JEE Main 2027 (Class 12)