The graph which represents the variation of m | vedclass

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

Question

(C) The mean kinetic energy $E$ of gas molecules is given by the formula:$E = \frac{f}{2} k T$where $f$ is the degrees of freedom,$k$ is the Boltzmann

Vedclass · Accepted Answer

Vedclass · Answer

(C) The mean kinetic energy $E$ of gas molecules is given by the formula:$E = \frac{f}{2} k T$where $f$ is the degrees of freedom,$k$ is the Boltzmann constant,and $T$ is the absolute temperature in Kelvin.Since $T = t + 273$,we can write:$E = \frac{f}{2} k (t + 273) = \left( \frac{f}{2} k ight) t + \left( \frac{f}{2} 	imes 273 k ight)$Comparing this with the equation of a straight line $y = mx + c$,where $y = E$ and $x = t$:Slope $m = \frac{f}{2} k$ (which is positive)Intercept $c = \frac{f}{2} 	imes 273 k$ (which is positive)Thus,the graph of $E$ versus $t$ is a straight line with a positive slope and a positive intercept on the $E$-axis. This corresponds to the graph shown in option $C$.

The graph which represents the variation of mean kinetic energy of molecules with temperature $t^{\circ} \text{C}$ is

Similar Questions

If the number of hydrogen molecules is twice the number of oxygen molecules at the same temperature,what is the ratio of their total kinetic energies? (Hydrogen : Oxygen)

If $r.m.s.$ velocity of a gas is $V_{rms} = 1840 \ m/s$ and its density $\rho = 8.99 \times 10^{-2} \ kg/m^3$,the pressure of the gas will be

Number of molecules in a volume of $4\, cm^{3}$ of a perfect monoatomic gas at some temperature $T$ and at a pressure of $2\, cm$ of mercury is close to $?$
(Given,mean kinetic energy of a molecule (at $T$) is $4 \times 10^{-14}\, erg$,$g=980\, cm/s^{2}$,density of mercury $=13.6\, g/cm^{3}$)

The translational kinetic energy of the molecules of a gas at absolute temperature $T$ can be doubled by

The kinetic energy,due to translational motion,of most of the molecules of an ideal gas at absolute temperature $T$ is proportional to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module