Question

The temperature of a gas is –78° C and the average translational kinetic energy of its molecules is K. The temperature at which the average translational kinetic energy of the molecules of the same gas becomes 2K is : 

(1) –39°C 

(2) 117°C 

(3) 127°C 

(4) –78°C 

Answer 1

Correct option is :  (2) 117°C

\(K . E=\frac{\mathrm{nf}_{1} \mathrm{RT}}{2}\)

\(\mathrm{T}_{\mathrm{i}}=-78^{\circ} \mathrm{C} \rightarrow 273+\left[-78^{\circ} \mathrm{C}\right]=195 \mathrm{K}\)

K.E \(\propto \ \mathrm{T}\)

To double the K.E energy temp also become double

\(\mathrm{T}_{\mathrm{f}}=390 \mathrm{~K}\)

\(\mathrm{T}_{\mathrm{f}}=117^{\circ} \mathrm{C}\)