Q62 / 90 · JEE Main 2012 (12 May Online) · Physics

A given ideal gas with $\gamma=\frac{C_p}{C_v}=1.5$ at a temperature $T$. If the gas is compressed adiabatically to one-fourth of its initial volume, the final temperature will be

  • A
    $2 \sqrt{2} T$
  • B
    $4 T$
  • Correct answer
    C
    $2 T$
  • D
    $8 T$

Solution

$T V^{\gamma-1}=$ constant $$ \begin{aligned} & T_1 V_1^{\gamma-1}=\mathrm{T}_2 V_2^{\gamma-1} \\ & \Rightarrow T(V)^{1 / 2}=\mathrm{T}_2\left(\frac{V}{4}\right)^{1 / 2} \\ & {\left[\because \gamma=1.5, T_1=T, V_1=V \text { and } V_2=\frac{V}{4}\right]} \\ & \therefore \quad T_2=\left(\frac{4 V}{V}\right)^{1 / 2} T=2 T \end{aligned} $$

Hence, the correct answer is option: 'C'.

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