Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice

Question

Two vessels separately contain two ideal gases A and B at the same temperature, the pressure of A being twice that of B. Under such conditions, the density of A is found to be 1.5 times the density of B. The ratio of molecular weight of A and B is:

Accepted Answer

According to the Ideal Gas Equation, $PV = nRT$. 
The number of moles $n$ can be written as $\frac{	ext{mass } (m)}{	ext{Molar Mass } (M)}$. 
Substituting this into the gas equation: $PV = \frac{m}{M}RT$. 
Rearranging to introduce density ($ho = \frac{m}{V}$): 
$P = \frac{m}{V} \frac{RT}{M} = ho \frac{RT}{M}$. 
Solving for Molar Mass ($M$): 
$M = \frac{ho RT}{P}$.

**Given conditions:** 
1. Temperature is constant ($T_A = T_B$). 
2. Pressure of A is twice that of B: $P_A = 2P_B \Rightarrow \frac{P_B}{P_A} = \frac{1}{2}$. 
3. Density of A is 1.5 times that of B: $ho_A = 1.5 ho_B \Rightarrow \frac{ho_A}{ho_B} = 1.5 = \frac{3}{2}$.

**Calculation:** 
Ratio of molecular weights $\frac{M_A}{M_B} = \frac{ho_A RT / P_A}{ho_B RT / P_B} = \left( \frac{ho_A}{ho_B} ight) 	imes \left( \frac{P_B}{P_A} ight)$. 
$\frac{M_A}{M_B} = \frac{3}{2} 	imes \frac{1}{2} = \frac{3}{4}$.

Step-by-Step Solution

Practice All PHYSICS Questions