Question

The r.m.s speed of molecules at a temperature 27 K and pressure 1.5 bar is $1\times {10}^4$cm/s. if both temperature and pressure are raised three times, the rms speed of the gas will be:

• A. $9\times {10}^{4}cm/s$
• B. $3\times {10}^{4}cm/s$
• C. $\sqrt{3}\times {10}^{4}cm/s$
• D. $1\times {10}^{4}cm/s$

Answer 1

The correct option is A $9\times {10}^{4}cm/s$

Given,

RMS of molecules $u_{rms}={10}^{4}cm/sec$

Pressure initially $=1.5bar$

Temperature $T_1=27K$

Now,we know

Temperature $T_2=2\times 27K$

Pressure $P_2=2\times 1.5bar$

Also,

$u_{rms}=\sqrt{\frac{3RT}{M}}$

It is evident that $u_{rms}$ is not affected by pressure.ONly change in temperature will affect.

Now,

$\frac{u_{rms1}}{u_{rms2}}=\frac{\sqrt{\frac{3R{T}_1}{M}}}{\sqrt{\frac{3R{T}_2}{M}}}$

$\frac{u_{rms1}}{u_{rms2}}=\sqrt{\frac{T_1}{T_2}}$

$\frac{{10}^4}{u_{rms2}}=\sqrt{\frac{27}{54}}$

$u_{rms2}=1.7\times {10}^{4}m/s$