Question

The temperature and pressure at Simla are 15.$0^0$ C and 72.0 cm of mercury and at Kalka these are 35.$0^0$ C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Simla.

Answer 1

For Simla

$T_1=15.0^0$ C = 288 K

$P_1$ = 72.0 cm of mercury

For Kalka

$T_2={35.0}^0$ C = 308 K

$P_2$ = 76.0 cm of mercury

From the formula

$PV=\frac{m}{M}$ RT

$\Rightarrow P=\left(\frac{m}{V}\right)\frac{1}{M}$ RT

$\Rightarrow P\frac{d}{M}$ RT

$\Rightarrow d=\frac{PM}{RT}$

Thus

$\frac{d^2}{d^1}=\frac{P_2}{T_2}\times \frac{T_1}{P_1}=\frac{76.0\times 288}{308\times 72.0}$

$\frac{d^2}{d^1}$ = .987