Question

The variation of translational kinetic energy of one mole of a monoatomic gas with temperature is shown in the graph. Find $\tan \theta$.

• A. $\frac{9R}{2}$
• B. $\frac{7R}{2}$
• C. $R$
• D. $\frac{3R}{2}$

Answer 1

The correct option is D $\frac{3R}{2}$

Given:
Number of moles, $n=1$
Nature of gas is monoatomic


We know that,
Average translational kinetic energy of an ideal gas is given by,
$KE=\frac{nfRT}{2}$ ............(1)
where,
$n\to$ number of moles
$f\to$ no of degrees of freedom
$R\to$ universal gas constant
$T\to$ temperature

On comparing (1) with the equation of straight line , $y=mx+c$,
where $y=KE$ & $x=T$
$\text{Slope},m=\tan \theta =\frac{nfR}{2}$
and $\text{y-intercept},c=0$

$\Rightarrow \tan \theta =\frac{3R}{2}$
[ for monoatomic gas, $f=3$ and given $n=1$ ]