Question

A gas sample containing 12 g of Ch4 has a volume of 14L. What will the volume be if 12 g of He is added?

Answer 1

Answer:

The volume of the gas sample after adding $12$ g of He is $70$ L.

Explanation:

Given that,

        $V_{1}$ $=$ $14$ L

       mass of $Ch_{4}$ in sample $=$ $12$ g

       mass of He added $=$ $12$ g

We know that,

        PV $=$ nRT

     $\frac{PV_{1} }{n_{1} RT}$  $=$  $\frac{PV_{2} }{n_{2}RT}$

Since we are adding He to the same sample, P and T are constants. Also, R is a constant value.

Then,

     $\frac{V_{1} }{n_{1} }$ $=$ $\frac{V_{2} }{n_{2} }$

     $V_{2}$ $=$ $\frac{V_{1} }{n_{1} }$ × $n_{2}$

To find out the no. of moles,

We know,

   n $=$ $\frac{Given mass}{Molar mass}$

⇒

  no. of. moles of the sample before He adding, $n_{1}$ $=$  $\frac{12}{16}$  $=$  $\frac{3}{4}$

  no. of moles of the sample after He added, $n_{2}$ $=$ $\frac{3}{4}$ $+$ $\frac{12}{4}$ $=$ $\frac{15}{4}$

∴      $V_{2}$ $=$ $\frac{14}{\frac{3}{4} }$  × $\frac{15}{4}$

           $=$ $14$ ×$5$

           $=$ $70$ L