Question

At constant temperature of $273K$, $\frac{1}{V}vsP$ are plotted for two ideal gases A and B as shown. Ratio of the number of moles of gas A and gas B are?

• A. $\frac{3}{1}$
• B. $\frac{1}{\sqrt{3}}$
• C. $\frac{1}{3}$
• D. $\frac{\sqrt{3}}{1}$

Answer 1

The correct option is D $\frac{\sqrt{3}}{1}$

We have the ideal gas equation, $PV=nRT$

Note that R is the ideal gas constant. T is also constant (given in the question).

n - number of moles of the gas.

We can rewrite this as, $P=\frac{1}{V}nRT$

$=>\frac{1}{V}=P\frac{1}{nRT}$ --------> ( 1 )

We know that equation of a straight line is , $y=mx+c$ ---------> ( 2 ) where, m is the slope.

The graph given in the question is plotted with $P$ in X-axis and $\frac{1}{V}$ in Y-axis.

So, by comparing equations ( 1 ) and ( 2 ), we get slope, $m=\frac{1}{nRT}$

For gas A

$n_A$ - number of moles of gas A

slope, $m=\frac{1}{n_{A}RT}$

We can calculate slope m as, $m=tan\theta$

where, $\theta$ - angle given in the graph

Given that, for gas A, $\theta ={45}^o$

So,

$tan\theta =\frac{1}{n_{A}RT}$

$=>tan{45}^o=\frac{1}{n_{A}RT}$

$=>1=\frac{1}{n_{A}RT}$

$=>n_{A}RT=1$ -----------> ( 3 )

For gas B

$n_B$ - number of moles of gas B

slope, $m=\frac{1}{n_{B}RT}$

Given that, for gas B, $\theta ={60}^o$

So,

$tan\theta =\frac{1}{n_{B}RT}$

$=>tan{60}^o=\frac{1}{n_{B}RT}$

$=>\sqrt{3}=\frac{1}{n_{B}RT}$

$=>n_{B}RT=\frac{1}{\sqrt{3}}$ -----------> ( 4 )

Divide equation ( 3 ) by ( 4 ),

$=>\frac{n_{A}RT}{n_{B}RT}=\frac{1}{\frac{1}{\sqrt{3}}}=\frac{\sqrt{3}}{1}$

$=>\frac{n_A}{n_B}=\frac{\sqrt{3}}{1}$

So, ratio of number of moles of gas A to gas B is $\frac{\sqrt{3}}{1}$