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Specific Heat Capacities of Gases — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsSpecific Heat Capacities of Gases5 Marks
Question
An ideal gas at pressure $2.5 \times 10^5Pa$ and temperature $300K$ occupies $100cc$. It is adiabatically compressed to half its original volume. Calculate,
  1. The final pressure.
  2. The final temperature.
  3. The work done by the gas in the process. Take $\gamma=1.5.$
✓

Answer

$P_1 = 2.5 \times 10^5Pa, V_1 = 100cc, T_1 = 300k$
  1. $\text{P}_1\text{V}_1^\gamma=\text{P}_2\text{V}_2^\gamma$
$\Rightarrow2.5\times10^5\times\text{V}^{1.5}=\Big(\frac{\text{V}}{2}\Big)^{1.5}\times\text{P}_2$
$\Rightarrow\text{P}_2=2^{1.5}\times2.5\times10^5$
$\Rightarrow\text{P}_2=7.07\times10^5\approx7.1\times10^5$
  1. $\text{T}_1\text{V}_1^{\gamma-1}=\text{T}_2\text{V}_2^{\gamma-1}$
$\Rightarrow300\times(100)^{1.5-1}=\text{T}_2\times(50)^{1.5-1}$
$\Rightarrow\text{T}_2=\frac{3000}{7.07}=424.32\text{k}\approx424\text{k}$
  1. Work done by the gas in the process,
$\text{W}=\frac{\text{mR}}{\gamma-1}[\text{T}_2-\text{T}_1]=\frac{\text{P}_1\text{V}_1}{\text{T}(\gamma-1)}[\text{T}_2-\text{T}_1]$
$\Rightarrow\frac{2.5\times10^5\times100\times10^{-6}}{300(1.5-1)}[424-300]$
$=\frac{2.5\times10}{300\times0.5}\times124=20.72\approx21\text{J}$

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