Kinetic Theory

JEE Main level, numerical answer, medium

6 of 168

Question 6 of 168

A gas of certain mass filled in a closed cylinder at a pressure of 3.23 kPa has temperature 50∘C50^{\circ} \mathrm{C}. The gas is now heated to double its temperature. The modified pressure is ____\_\_\_\_ Pa. Note: Volume is constant.

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Answer: 3730

Since its given that volume is constant, P∝TP \propto T.

Ti=50°C=323T_i = 50°C = 323 K
Tf=2×50=100°C=373T_f = 2 \times 50 = 100°C = 373 K

PfPi=TfTi\frac{P_f}{P_i} = \frac{T_f}{T_i}

Pf3.23 kPa=373323\frac{P_f}{3.23 \text{ kPa}} = \frac{373}{323}

Pf=3.23×373323=3.73P_f = 3.23 \times \frac{373}{323} = 3.73 kPa =3730= 3730 Pa

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