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$.

$X = x$	$0$	$1$	$2$	$3$
$P(X = x)$	$\frac{1}{3}$	$\frac{1}{2}$	$0$	$\frac{1}{6}$

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

Which of the following statements is incorrect for $Nostoc$ and $Anabaena$?

What is the time (in $sec$) required for depositing all the silver present in $125 \ mL$ of $1 \ M \ AgNO_3$ solution by passing a current of $241.25 \ A$? $(1 \ F = 96500 \ C)$

If $f(x) = \frac{x^2}{x+a}$, then $f^{\prime \prime}(a)$ is equal to

$\sinh^{-1}(2^{3/2})$ is equal to :

If $m$ and $\sigma^2$ are the mean and variance of the random variable $X$,whose distribution is given by:
$X = x$ $0$ $1$ $2$ $3$
$P(X = x)$ $\frac{1}{3}$ $\frac{1}{2}$ $0$ $\frac{1}{6}$

Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module