The temperature of a gas having 2 . 0 × 10 25 molecules per cubic meter at 1 . 38 atm (Given, k = 1 . 38 × 10 - 23 J K - 1 ) is :

From ideal gas equation, P V = n R T ⇒ P V = N N A R T , where N = total number of molecules ⇒ P = N V R N A T ⇒ P = N V k T ⇒ 1 . 38 × 1 . 01 × 10 5 = 2 × 10 25 × 1 . 38 × 10 - 23 × T ⇒ 1 . 01 × 10 5 = 2 × 10 2 × T T = 1 . 01 × 10 3 2 ≈ 500 K

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The temperature of a gas having $2.0 \times 10^{25}$ molecules per cubic meter at $1.38 atm$ (Given, $k = 1.38 \times 10^{- 23} J K^{- 1})$ is :

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From ideal gas equation, $P V = n R T$

$\Rightarrow P V = \frac{N}{N_{A}} R T$ , where $N =$ total number of molecules

$\Rightarrow P = \frac{N}{V} (\frac{R}{N_{A}}) T \Rightarrow P = \frac{N}{V} k T$

$\Rightarrow 1.38 \times (1.01 \times 10^{5}) = (2 \times 10^{25}) \times (1.38 \times 10^{- 23}) \times T$

$\Rightarrow 1.01 \times 10^{5} = 2 \times 10^{2} \times T$

$T = \frac{1.01 \times 10^{3}}{2} \approx 500 K$

Stream:JEESubject:PhysicsTopic:Heat & ThermoSubtopic:KTG

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