Question

The temperature of an ideal gas is increased from 27${}^o$C to 927${}^o$C. Then, the mean kinetic energy of gas molecules :

• A. becomes $\frac{927}{27}$ times
• B. becomes $\sqrt{\frac{927}{27}}$ times
• C. become double
• D. become four times

Answer 1

The correct option is D become four times

$T_1={27}^{0}C=300K$

$T_2={927}^{0}C=1200K$

$V_{rms}$( root mean square velocity )$=\sqrt{\frac{3RT}{M}}$

$K{E}_1=\frac{1}{2}\times m\times V_{rms}^2$

$K{E}_2=\frac{1}{2}\times m\times \frac{3R\times 1200}{m}$

$=4K{E}_1$