An ideal gas is found to obey $\mathrm{Pv}^{\frac{3}{2}}=$ constant during an adiabatic process. If such a…

An ideal gas is found to obey $\mathrm{Pv}^{\frac{3}{2}}=$ constant during an adiabatic process. If such a gas initially at a temperature T is aḍiabatically compressed to $\frac{1}{4}$ th of its volume, then its final temperature is
  1. $\sqrt{3 \mathrm{~T}}$
  2. $\sqrt{2 \mathrm{~T}}$
  3. 2 T
  4. 3 T

Solution

$\mathrm{Pv}^{\frac{3}{2}}=$ constant...(i)
$\text { Also, } \mathrm{PV}=\mathrm{nRT} \Rightarrow \mathrm{p}=\frac{\mathrm{nRT}}{\mathrm{~V}}$...(ii)
From eq(i) and (ii), we get $\left(\frac{\mathrm{nRT}}{\mathrm{~V}}\right) \mathrm{V}^{\frac{3}{2}}=\mathrm{k} \Rightarrow T \mathrm{~V}^{\frac{1}{2}}=\mathrm{c}$
Now, $V_2=\frac{V_1}{4}$ $\therefore \mathrm{T}_1 \mathrm{~V}_1^{\frac{1}{2}}=\mathrm{T}_2 \mathrm{~V}_2^{\frac{1}{2}} \Rightarrow \mathrm{TV}_1^{\frac{1}{2}}=\mathrm{T}_2\left(\frac{\mathrm{~V}_1}{4}\right)^{\frac{1}{2}}$ $\therefore \quad \mathrm{T}_2=2 \mathrm{~T}$

Asked in: AP EAMCET 2024 (21 May Shift 2)

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