Kinetic Theory

JEE Main level, single correct, medium

10 of 168

Question 10 of 168

The temperature of a gas having 2.0×10252.0\times {10}^{25} molecules per cubic meter at 1.38atm1.38atm (Given, k=1.38×10−23JK−1)k=1.38\times {10}^{-23}J{K}^{-1}) is :

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Answer: (A) 500K500K

From ideal gas equation, PV=nRTPV=nRT
⇒PV=NNART\Rightarrow PV=\dfrac{N}{{N}_{A}}RT, where N=N= total number of molecules
⇒P=NV(RNA)T⇒P=NVkT\Rightarrow P=\dfrac{N}{V}\left(\dfrac{R}{{N}_{A}}\right)T \Rightarrow P=\dfrac{N}{V}kT
⇒1.38×(1.01×105)=(2×1025)×(1.38×10−23)×T\Rightarrow 1.38\times \left(1.01\times {10}^{5}\right)=\left(2\times {10}^{25}\right)\times \left(1.38\times {10}^{-23}\right)\times T
⇒1.01×105=2×102×T\Rightarrow 1.01\times {10}^{5}=2\times {10}^{2}\times T
T=1.01×1032≈500KT=\dfrac{1.01\times {10}^{3}}{2}\approx 500K

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