When the temperature of a gas is increased fr | vedclass

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

Question

(A) The root mean square speed $(u_{rms})$ is directly proportional to the square root of the absolute temperature $(T)$: $u_{rms} = \sqrt{\frac{3RT}{

Vedclass · Accepted Answer

get doubled

Vedclass · Answer

(A) The root mean square speed $(u_{rms})$ is directly proportional to the square root of the absolute temperature $(T)$: $u_{rms} = \sqrt{\frac{3RT}{M}}$ For the same gas,$u_{rms} \propto \sqrt{T}$. Initial temperature $T_{1} = 30 + 273 = 303 \ K$. Final temperature $T_{2} = 930 + 273 = 1203 \ K$. Taking the ratio: $\frac{u_{2}}{u_{1}} = \sqrt{\frac{T_{2}}{T_{1}}} = \sqrt{\frac{1203}{303}} = \sqrt{3.97} \approx \sqrt{4} = 2$. Therefore,$u_{2} = 2u_{1}$. Thus,the root mean square speed of the gas gets doubled.

When the temperature of a gas is increased from $30^{\circ} C$ to $930^{\circ} C$,the root mean square speed of the gas would

Similar Questions

If the average velocity of $N_2$ gas at $27 \, ^oC$ is $0.3 \, m/sec$,then at what temperature in $K$ will the average velocity be $0.6 \, m/sec$?

The molecular weight of a gas that diffuses twice as rapidly as the gas with molecular weight $64$ is:

$A$ gas bulb of $1 \ L$ capacity contains $2.0 \times 10^{21}$ molecules of nitrogen gas at a pressure of $7.57 \times 10^3 \ N \ m^{-2}$. Calculate the root mean square speed in $cm \ sec^{-1}$. (in $.5$)

The volumes of ozone and chlorine gas diffusing in the same time are $35 \ mL$ and $29 \ mL$ respectively. If the molar mass of chlorine is $71$,calculate the molar mass of ozone. (in $.7$)

The $rms$ velocity of hydrogen is $\sqrt{7}$ times the $rms$ velocity of nitrogen. If the temperature of the gas is $T$,then which of the following is true?

Vedclass Test Series

Exam Paper Generator

Online Exam Module