Question:
A monoatomic ideal gas is compressed adiabatically to \(\left(\frac{1}{27}\right)\) of its initial volume. If initial temperature of the gas is ' T ' K and final temperature is ' xT ' K , the value of ' x ' is
A monoatomic ideal gas is compressed adiabatically to \(\left(\frac{1}{27}\right)\) of its initial volume. If initial temperature of the gas is ' T ' K and final temperature is ' xT ' K , the value of ' x ' is
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For an adiabatic process of an ideal gas, T times V raised to gamma minus one is constant. Use gamma for a monoatomic gas.
Updated On: Nov 13, 2026
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
For an adiabatic process, \(TV^{\gamma - 1} = \text{constant}\).
For a monoatomic ideal gas \(\gamma = \dfrac{5}{3}\), so \(\gamma - 1 = \dfrac{2}{3}\).
Step 2: Substitution
\(T_1V_1^{2/3} = T_2V_2^{2/3}\) with \(V_2 = \dfrac{V_1}{27}\).
Step 3: Calculation
\(T_2 = T\left(\dfrac{V_1}{V_2}\right)^{2/3} = T\,(27)^{2/3} = 9T\).
So \(x = 9\).
Final Answer:
The value of x is 9. \[ \boxed{\text{(B) }9} \]
For an adiabatic process, \(TV^{\gamma - 1} = \text{constant}\).
For a monoatomic ideal gas \(\gamma = \dfrac{5}{3}\), so \(\gamma - 1 = \dfrac{2}{3}\).
Step 2: Substitution
\(T_1V_1^{2/3} = T_2V_2^{2/3}\) with \(V_2 = \dfrac{V_1}{27}\).
Step 3: Calculation
\(T_2 = T\left(\dfrac{V_1}{V_2}\right)^{2/3} = T\,(27)^{2/3} = 9T\).
So \(x = 9\).
Final Answer:
The value of x is 9. \[ \boxed{\text{(B) }9} \]
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