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(C) The root mean square speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$,we have the ratio $\frac{(v_{rms

Vedclass · Accepted Answer

It increases by $0.732$ times

Vedclass · Answer

(C) The root mean square speed of a gas is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$.Since $v_{rms} \propto \sqrt{T}$,we have the ratio $\frac{(v_{rms})_{2}}{(v_{rms})_{1}} = \sqrt{\frac{T_{2}}{T_{1}}}$.Given that the temperature is increased to $3$ times,$T_{2} = 3T_{1}$.Substituting this into the ratio,we get $\frac{(v_{rms})_{2}}{(v_{rms})_{1}} = \sqrt{\frac{3T_{1}}{T_{1}}} = \sqrt{3}$.Therefore,$(v_{rms})_{2} = \sqrt{3}(v_{rms})_{1} \approx 1.732(v_{rms})_{1}$.The change in $v_{rms}$ is $(v_{rms})_{2} - (v_{rms})_{1} = 1.732(v_{rms})_{1} - 1(v_{rms})_{1} = 0.732(v_{rms})_{1}$.Thus,the $v_{rms}$ increases by $0.732$ times the initial value.

When the temperature of a gas is increased to $3$ times,what will be the change in $v_{rms}$?

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