Kinetic Theory

JEE Main level, single correct, medium

141 of 168

Question 141 of 168

A cylindrical container of volume 4.0×10−3m34.0\times {10}^{-3}{m}^{3} contains one mole of hydrogen and two moles of carbon dioxide. Assume the temperature of the mixture is 400K400K. The pressure of the mixture of gases is :
[Take gas constant as 8.3Jmol−1K−1]\left.8.3J{mol}^{-1}{K}^{-1}\right]

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Answer: (C) 24.9×105Pa24.9\times {10}^{5}Pa

For gas A
PAV=nART{P}_{A}V={n}_{A}RT
PA=(1)RTV{P}_{A}=\dfrac{(1)RT}{V}
for gas BB
PBV=nBRT{P}_{B}V={n}_{B}RT
PB=(2)RTV{P}_{B}=\dfrac{(2)RT}{V}
Now according to the Dalton's law of mixture,
P=PA+PBP={P}_{A}+{P}_{B}
=3RTV=3×8.314(400)4×10−3=24.9×105Pa=\dfrac{3RT}{V}=\dfrac{3\times 8.314(400)}{4\times {10}^{-3}}=24.9\times {10}^{5}Pa

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