Question

A balloon containing 1 mol of air initiallyat 1 atm is filled further with air till the pressure increases to 4 atm. The initial diameter of the balloon is $1\text{m}$ and the pressure at each stage is proportional to diameter of the balloon. How many moles of air was to change the pressure from 1 atm to 4 atm?

• A. $80$
• B. $257$
• C. $255$
• D. $256$

Answer 1

The correct option is C $255$

$n_1=1\text{mole},n_2=?,P_1=1\text{atm},P_2=4\text{atm}$
$P_1{V}_1=n_{1}RT$
$P_2{V}_2=n_{2}RT$
$\frac{P_1{V}_1}{n_1}=\frac{P_2{V}_2}{n_2}$
$\Rightarrow \frac{1}{1}\times \frac{4}{3}\pi (r_1{)}^3=\frac{4}{n_2}\times \frac{4}{3}\pi (r_2{)}^3$
$\Rightarrow (r_1{)}^3=\frac{4}{n_2}(r_2{)}^3...\left(i\right)$
Now, $r_1=\frac{1}{2}\text{m (given)};r_2=?$
But, $P\propto d\text{(given)}\Rightarrow \frac{P_1}{d_1}=\frac{P_2}{d_2}\Rightarrow \frac{1}{1}=\frac{4}{d_2}$
$d_2=4\text{m}\Rightarrow r_2=2\text{m}...\left(ii\right)$
$\Rightarrow$ From (ii) and (i), we have
${\left(\frac{1}{2}\right)}^3=\frac{4}{n_2}(2{)}^3$
$n_2=64\times 4$
$n_2=256$
$\Rightarrow$ Number of moles added $=256-1=255$