Question

The temperature and pressure at Simla are 15.0°C and 72.0 cm of mercury and at Kalka these are 35.0°C and 76.0 cm of mercury. Find the ratio of air density at Kalka to the air density at Simla.

Answer 1

The ratio of air density at Kalka to the air density at Simla is 0.986.

• To find the density we use the relation P = ρ * R * T

• So, calculating the pressure at Shimla P1 = 0.72 * 13600 * 9.81 = 96059.52 Pa
• Calculating the pressure at Kalka P2 = 0.76 * 13600 * 9.81 = 101396.16 Pa
• Now, calculating the density of air at shimla = 96059.52/(287* 288) = 1.162 Kg/m^3
• Calculating the density of air at Kalka = 101396.16/ (287*308) =1.147 Kg/m^3

Finally the ratio of the densities is 0.986.

Answer 2

The ratio of air density at Kalka to the air density at Simla is $1.013$

Explanation:

Step 1:

Given data,  

Simla’s  Temperature , $T_{1}=15+273=288 \mathrm{K}$

Simla’s Pressure , $\mathrm{P}_{1}=0.72 \mathrm{m} \text { of } \mathrm{Hg}$

Kalka’s  Temperature , $T_{2}=35+273=308 \mathrm{K}$

Kalka’s Pressure , $\mathrm{P}_{2}=0.76 \mathrm{m} \text { of } \mathrm{Hg}$

Let air density be at Simla and Kalka

$\rho_{1}$ and

$\rho_{2}$, respectively.  

Step 2:

Then,

We know that

$P V=\frac{m R T}{M}$

$\frac{m}{V}=\frac{P M}{R T}$

$\rho=\frac{P M}{R T}$

thus  

$\rho_{1}=\frac{P_{1} M}{R T_{1}}$

$\rho_{2}=\frac{P_{2} M}{R T_{2}}$

Step 3:

We take ratios

$\frac{\rho_{1}}{\rho_{2}}=\frac{P_{1}}{T_{1}} \times \frac{T_{2}}{P_{2}}$

$\frac{\rho_{1}}{\rho_{2}}=\frac{0.72}{288} \times \frac{308}{0.76}$

$\frac{\rho_{1}}{\rho_{2}}=2.5 \times 10^{-3} \times 405.26$

$\frac{\rho_{1}}{\rho_{2}}=1.013$