JEE Mains · Physics · STD 11 - 10.1, thermonetry,thermal expansion and calorimetry
- JEE Mains 2026
- Medium
- MCQ
An ideal gas at pressure \(P\) and temperature \(T\) is expanding such that \(PT^3 = \) constant. The coefficient of volume expansion of the gas is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(\dfrac{4}{T}\)
Step-by-step Solution
Detailed explanation
Given the process equation for the ideal gas:
\(PT^3 = \text{constant}\)
From the ideal gas equation, \(PV = nRT\), which gives \(P = \dfrac{nRT}{V}\).
Substituting \(P\) into the process equation:
\(\left(\dfrac{nRT}{V}\right) T^3 = \text{constant}\)
\(\dfrac{T^4}{V} = \text{constant}\)
\(V = k T^4\)
Differentiating both sides with respect to temperature \(T\):
\(\dfrac{dV}{dT} = 4k T^3\)
The coefficient of volume expansion \(\gamma\) is defined as:
\(\gamma = \dfrac{1}{V} \dfrac{dV}{dT}\)
Substituting the values of \(V\) and \(\dfrac{dV}{dT}\):
\(\gamma = \dfrac{1}{k T^4} (4k T^3) = \dfrac{4}{T}\)
Answer: \(\dfrac{4}{T}\)
\(PT^3 = \text{constant}\)
From the ideal gas equation, \(PV = nRT\), which gives \(P = \dfrac{nRT}{V}\).
Substituting \(P\) into the process equation:
\(\left(\dfrac{nRT}{V}\right) T^3 = \text{constant}\)
\(\dfrac{T^4}{V} = \text{constant}\)
\(V = k T^4\)
Differentiating both sides with respect to temperature \(T\):
\(\dfrac{dV}{dT} = 4k T^3\)
The coefficient of volume expansion \(\gamma\) is defined as:
\(\gamma = \dfrac{1}{V} \dfrac{dV}{dT}\)
Substituting the values of \(V\) and \(\dfrac{dV}{dT}\):
\(\gamma = \dfrac{1}{k T^4} (4k T^3) = \dfrac{4}{T}\)
Answer: \(\dfrac{4}{T}\)
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