The temperature of a gas is -73^{circ}C. To w | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The initial temperature $T_1 = -73^{\circ}C = (-73 + 273) \, K = 200 \, K$.The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT

Vedclass · Accepted Answer

$527$

Vedclass · Answer

(D) The initial temperature $T_1 = -73^{\circ}C = (-73 + 273) \, K = 200 \, K$.The $rms$ speed of gas molecules is given by $v_{rms} = \sqrt{\frac{3RT}{M}}$,which implies $v_{rms} \propto \sqrt{T}$.Let the initial $rms$ speed be $v$ and the final $rms$ speed be $2v$.Using the ratio: $\frac{v_{rms,2}}{v_{rms,1}} = \sqrt{\frac{T_2}{T_1}}$.Substituting the values: $\frac{2v}{v} = \sqrt{\frac{T_2}{200}}$.Squaring both sides: $4 = \frac{T_2}{200}$.Therefore,$T_2 = 800 \, K$.Converting back to Celsius: $T_2(^{\circ}C) = 800 - 273 = 527^{\circ}C$.

The temperature of a gas is $-73^{\circ}C$. To what temperature should the gas be heated so that the $rms$ speed of the molecules is doubled (in $^{\circ}C$)?

Similar Questions

For an ideal gas with constant pressure,the root mean square velocity $v_{rms}$ is proportional to . . . . . . .

When the temperature of a gas is raised from $27^{\circ} C$ to $90^{\circ} C$,the increase in the rms velocity of the gas molecules is: (in $\%$)

The absolute temperature of a gas is determined by

At what temperature in $K$ will the $r.m.s.$ speed of a hydrogen molecule be equal to the escape velocity from the Earth?

The given diagram shows isotherms for a fixed mass of an ideal gas at temperatures $T_1$ and $T_2$. What is the value of the ratio $\frac{\text{r.m.s. speed of the molecules at temperature } T_2}{\text{r.m.s. speed of the molecules at temperature } T_1}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module