Question:
An ideal rigid diatomic gas \((γ = \frac{7}{5})\) undergoes an adiabatic change . If the relation between temperature and volume is \(TV^x =\) constant then the value of x is
An ideal rigid diatomic gas \((γ = \frac{7}{5})\) undergoes an adiabatic change . If the relation between temperature and volume is \(TV^x =\) constant then the value of x is
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x equals gamma minus 1.
Updated On: Oct 1, 2026
- \(0.25\)
- \(0.30\)
- \(0.40\)
- \(0.60\)
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept:
For an ideal gas undergoing an adiabatic change, \(PV^\gamma = \text{constant}\). Using \(P = \frac{nRT}{V}\) we get \(TV^{\gamma-1} = \text{constant}\).
Step 2: Compute:
For a rigid diatomic gas, \(\gamma = \frac75\). So
\[ x = \gamma - 1 = \frac75 - 1 = \frac25 = 0.4 \]
Step 3: Why the other options are wrong.
The value 0.25 would result from using \(\gamma = 1.25\). Values 0.30 and 0.60 do not match \(\gamma - 1\) for \(\gamma = 1.4\).
Final Answer:
The value of x is \(0.40\), option (C). \[ \boxed{0.40} \]
For an ideal gas undergoing an adiabatic change, \(PV^\gamma = \text{constant}\). Using \(P = \frac{nRT}{V}\) we get \(TV^{\gamma-1} = \text{constant}\).
Step 2: Compute:
For a rigid diatomic gas, \(\gamma = \frac75\). So
\[ x = \gamma - 1 = \frac75 - 1 = \frac25 = 0.4 \]
Step 3: Why the other options are wrong.
The value 0.25 would result from using \(\gamma = 1.25\). Values 0.30 and 0.60 do not match \(\gamma - 1\) for \(\gamma = 1.4\).
Final Answer:
The value of x is \(0.40\), option (C). \[ \boxed{0.40} \]
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