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Physics Question on Kinetic molecular theory of gases

The average kinetic energy of a gas molecules at 27∘C27^{\circ} C is 6.21×10−21J6.21 \times 10^{-21} J. Its average kinetic energy at 227∘C227^{\circ}C will be

A

10.35×10−21J10.35 \times 10^{-21} J

B

52.2×10−21J52.2 \times 10^{-21} J

C

5.22×10−21J5.22 \times 10^{-21} J

D

11.35×10−21J11.35 \times 10^{-21} J

Answer

10.35×10−21J10.35 \times 10^{-21} J

Explanation

Solution

We know that the K.EK.E of one mole of a gas molecules at a temperature TT is given by K=32KTK =\frac{3}{2} K T Now,T1=27∘C=300 KT_{1}=27^{\circ} C=300\, K K1=6.21×10−21 JK_{1}=6.21 \times 10^{-21}\, J T2=227∘C=500 KT_{2}=227^{\circ} C=500\, K K2=?K_{2}=? We have, K1K2=T1T2\frac{K_{1}}{K_{2}}=\frac{T_{1}}{T_{2}} ⇒K2=T1T2×K1=500300×6.21×10−21 \Rightarrow K_{2}=\frac{T_{1}}{T_{2}} \times K_{1}=\frac{500}{300} \times 6.21 \times 10^{-21} =10.35×10−21 J=10.35 \times 10^{-21}\, J