The average kinetic energy of a gas molecule | vedclass

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

Question

(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{

Vedclass · Accepted Answer

$10.35 	imes 10^{-21}\,J$

Vedclass · Answer

(C) The average kinetic energy $(E)$ of a gas molecule is directly proportional to its absolute temperature $(T)$,given by the relation $E = \frac{3}{2} k_B T$,where $k_B$ is the Boltzmann constant.Therefore,$E \propto T$,which implies $\frac{E_1}{E_2} = \frac{T_1}{T_2}$.Given temperatures in Kelvin:$T_1 = 27 + 273 = 300\,K$$T_2 = 227 + 273 = 500\,K$Given energy $E_1 = 6.21 	imes 10^{-21}\,J$.Substituting the values into the ratio:$\frac{6.21 	imes 10^{-21}}{E_2} = \frac{300}{500}$$\frac{6.21 	imes 10^{-21}}{E_2} = \frac{3}{5}$$E_2 = \frac{6.21 	imes 10^{-21} 	imes 5}{3}$$E_2 = 2.07 	imes 10^{-21} 	imes 5 = 10.35 	imes 10^{-21}\,J$.

The average kinetic energy of a gas molecule at $27\,^{\circ}C$ is $6.21 \times 10^{-21}\,J$. Its average kinetic energy at $227\,^{\circ}C$ will be

Similar Questions

If the rate of disappearance of $N_2O_5$ in the following reaction is $1.2 \times 10^{-5} \ mol \ L^{-1} \ s^{-1}$,the rate of production of $NO_2$ in $mol \ L^{-1} \ s^{-1}$ is:
$2 \ N_2O_{5(g)} \longrightarrow 4 \ NO_{2(g)} + O_{2(g)}$

$A$ gaseous mixture contains $50\%$ helium and $50\%$ methane by volume. Find the percentage by weight of methane in this gaseous mixture.

Which of the following reagents will convert propanoic acid to propan-$1$-ol?

The diameter of the objective of a telescope is $1 \ m$. Its resolving limit for light of wavelength $4538 \ \mathring{A}$ will be:

Assertion: The isothermal curves intersect each other at a certain point.
Reason: The isothermal change takes place slowly,so,the isothermal curves have very little slope.

Vedclass Test Series

Exam Paper Generator

Online Exam Module