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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Physics
Chapter
Thermodynamics

BITSAT 2023 · Paper 2 · Physics · Thermodynamics

A hypothetical gas expands adiabatically such that its volume...

A hypothetical gas expands adiabatically such that its volume changes from 08 litres to 27 litres. If the ratio of final pressure of the gas to initial pressure of the gas is 1681\frac{16}{81}. Then the ratio of CPCV\frac{\mathrm{C}_{\mathrm{P}}}{\mathrm{C}_{\mathrm{V}}} will be.

Options

  1. A43\frac{4}{3}
  2. B31\frac{3}{1}
  3. C12\frac{1}{2}
  4. D32\frac{3}{2}

Correct answer

A. 43\frac{4}{3}

Step-by-step solution

In adiabatic processPVγ= constant ⇒P1V1γ=P2V2γ⇒P1P2=(V2V1)γ⇒8116=(278)γ⇒(94)2=(9×34×2)γ⇒(94)2−γ=(32)γ⇒(94)2−γ=(94)γ/2⇒2−γ=γ2⇒2=32γ⇒γ=43\begin{aligned} \\ & \mathrm{PV}^\gamma=\text{ constant } \\ \\ & \Rightarrow \quad \mathrm{P}_1 \mathrm{ V}_1^\gamma=\mathrm{P}_2 \mathrm{ V}_2^\gamma \\ \\ & \Rightarrow \quad \frac{\mathrm{P}_1}{\mathrm{P}_2}=\left(\frac{\mathrm{V}_2}{\mathrm{ V}_1}\right)^\gamma \Rightarrow \frac{81}{16}=\left(\frac{27}{8}\right)^\gamma \\ \\ & \Rightarrow\left(\frac{9}{4}\right)^2=\left(\frac{9 \times 3}{4 \times 2}\right)^\gamma \Rightarrow\left(\frac{9}{4}\right)^{2-\gamma}=\left(\frac{3}{2}\right)^\gamma \\ \\ & \Rightarrow\left(\frac{9}{4}\right)^{2-\gamma}=\left(\frac{9}{4}\right)^{\gamma / 2} \Rightarrow 2-\gamma=\frac{\gamma}{2} \\ \\ & \Rightarrow 2=\frac{3}{2} \gamma \Rightarrow \gamma=\frac{4}{3} \\ \end{aligned}
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