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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Kinetic Theory of Gases
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For an ideal gas, the following conversions are made:

$1) P, V, T (\text{initial conditions}) \;\;\underrightarrow{\text{Isothermally}} \;\;2V, P_i, T$

$2) P, V, T (\text{initial conditions}) \;\;\underrightarrow{\text{Adiabatically}} \;\;2V_i, P_a,T$

Now, if $K_{gas} = \large\frac{P_a}{P_i}$, find $\large \frac{K_{H_2}}{K_{He}}$ (A) 1 (B) 1.2 (C) 0.83 (D) 1.44
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1 Answer

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For any ideal gas, $P_i \times 2V = P \times V \rightarrow P_i = \large \frac{1}{2}$$P$
For an ideal gas expanded adiabatically, $P_a (2V)^{\lambda} = PV^{\lambda} \rightarrow P_a = \large\frac{1}{2^{\lambda}}$$P$
$\Rightarrow \large\frac{P_a}{P_i} $$=\large\frac{2}{2^{\lambda}}$
For $H_2,\; \lambda = \large\frac{7}{5}$ $ \rightarrow K_{H_{2}} = \large\frac{2}{2^{1.4}}$$ = 0.758$
For $He,\; \lambda = \large\frac{5}{3}$ $ \rightarrow K_{He} = \large\frac{2}{2^{1.666}}$$ = 0.630$
$\Rightarrow \large \frac{K_{H_2}}{K_{He}}$$ \approx 1.2$
answered Mar 11, 2014 by balaji.thirumalai
 

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