JEE Advanced · Physics · 13. Thermodynamics
An ideal gas is expanding such that \(p T^2=\) constant. The coefficient of volume expansion of the gas is
- A \(\frac{1}{T}\)
- B \(\frac{2}{T}\)
- C \(\frac{3}{T}\)
- D \(\frac{4}{T}\)
Answer & Solution
Correct Answer
(C) \(\frac{3}{T}\)
Step-by-step Solution
Detailed explanation
\(p T^2=\) constant
\(\therefore \left(\frac{n R T}{V}\right) T^2=\text { constant } \)
\( \text { or } T^3 V^{-1}=\text { constant }\)
Differentiating the equation, we get
\(
\frac{3 T^2}{V} \cdot d T-\frac{T^3}{V^2} \cdot d V=0
\)
or
\(
3 \cdot d T=\frac{T}{V} \cdot d V
\)
From the equation, \(d V=V \gamma \cdot d T\)
\(\gamma\) = coefficient of volume expansion of gas \(=\frac{d V}{V \cdot d T}\)
From Eq. (i) \(\gamma=\frac{d V}{V \cdot d T}=\frac{3}{T}\)
\(\therefore\) correct answer is (c).
\(\therefore \left(\frac{n R T}{V}\right) T^2=\text { constant } \)
\( \text { or } T^3 V^{-1}=\text { constant }\)
Differentiating the equation, we get
\(
\frac{3 T^2}{V} \cdot d T-\frac{T^3}{V^2} \cdot d V=0
\)
or
\(
3 \cdot d T=\frac{T}{V} \cdot d V
\)
From the equation, \(d V=V \gamma \cdot d T\)
\(\gamma\) = coefficient of volume expansion of gas \(=\frac{d V}{V \cdot d T}\)
From Eq. (i) \(\gamma=\frac{d V}{V \cdot d T}=\frac{3}{T}\)
\(\therefore\) correct answer is (c).
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