Question

The pressure exerted by 12 g of an ideal gas at temperature $t{}^{\circ }C$ in a vessel of volume $V$ litre is 1 atm. When the temperature is increased by $10{}^{\circ }C$ at the same volume, the pressure increases by 10%. Calculate the volume of $V$.

[Molecular weight of the gas is 120. Write the value of V $\times$ 1000 Value]

Answer 1

$PV=nRT$

$PV=nRT=\frac{w}{m}RT$

$1\times V=\frac{12}{120}\times R\times (273+t)$ (i)

$1.1\times V=\frac{12}{120}\times R\times (273+t+10)$ (ii)

Dividing equation (ii) by equation (i), we get

$1.1=\frac{(273+t+10)}{(273+t)}$

or $t=-173{}^{\circ }C$

Putting this value in equation (i), we get

$1\times V=\frac{12}{120}\times 0.0821\times (273-173)$

or $V=0.821L$

$\therefore V=x\times 1000=0.821\times 1000=821$