For VT vs T curve, for 1 mol of gas find pressure of gas in atm if [ dT d ( VT ) ] 300 K = 24.63 .

Charle's law, V = K T d T d ( V T ) = d T d ( K T 2 ) = 2 K T = P 2 2 n R × T ⇒ 24.63 = P 2 × 1 × 0.0821 × 300 ⇒ P = 2 a t m

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ChemistryIdeal GasIdeal Gas Equation & problemsMedium2 minQB

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For VT vs T curve, for 1 mol of gas find pressure of gas in atm if $[\frac{d ( VT )}{dT}]_{300 K} = 24.63$ .

Answer:

2.00

Solution:

Charle's law, $V = K T$

$\frac{d ( V T )}{d T} = \frac{d}{d T} (K T^{2}) = 2 K T = \frac{2 n R}{P ^{2}} \times T$

$\Rightarrow 24.63 = \frac{2 \times 1 \times 0.0821}{P} \times 300$

$\Rightarrow P = 2 a t m$

Stream:JEESubject:ChemistryTopic:Ideal GasSubtopic:Ideal Gas Equation & problems

⏱ 2mℹ️ Source: QB

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