Question

A compound exists in the gaseous phase both as monomer(A) and dimer(B). The molecular mass of A is 48. In an experiment, $96g$ of the compound was confined in a vessel of volume $33.61$ and heated to $546K$ . What is the pressure developed if the compound exists as a dimer to the extent of $50\% $ by weight under these conditions?
A) $2atm$
B) $4atm$
C) $3atm$
D) $0.5atm$

Answer 1

Hint: The ideal gas equation is given as: $PV = nRT$
Where, P is the pressure (in atm), V is the volume (in Litres) , n is the total moles of the gas present in the mixture, R is the gas constant and T is the temperature (in Kelvin). The value of R changes with the given Pressure and Volume. The various values of R can be given as: $8.314J/K/mol,0.0821Latm/mol,2cal/Kmol$. This equation is not applicable to Real Gases. If the nature of the gas is not specified, consider it as ideal only.

Complete answer:
The information provided to us is:
Mass of dimer (B) is $50\% $ to the mixture of the mixture of both Monomer and Dimer
Molecular mass of Dimer A $ = 48g/mol$
Volume of the vessel $ = 33.61L$
Temperature $ = 546K$
Mixture of the compound used $ = 96g$
If we are given $100g$ of the compound then $50g$ would be dimer (B). Therefore, in $96g$ of mixture we will have $50\% \times 96 = 48g$ of Dimer (B).
The mass of Monomer A will be = Total Mass of the Compound – Mass of Dimer B
Mass of (A) will be $ = 96 - 48 = 48g$ of Monomer A
Finding the number of moles of Monomer A and Dimer B;
No. of moles of Monomer A $ = \dfrac{{48}}{{48}} = 1mol$
No. of moles of Dimer B $ = \dfrac{{48}}{{96}} = 0.5mol$
Total no. of moles of both A and B $ = 1 + 0.5 = 1.5mol$
To find the pressure, we’ll use the ideal gas equation. The pressure can be given as:
$P = \dfrac{{nRT}}{V} = \dfrac{{1.5 \times 0.0821 \times 546}}{{33.61}}$
$P = 2atm$

The correct option is Option (A).

Note:
This problem can be alternatively solved by using the formula; $PV = ({n_1} + {n_2})RT$
Where, ${n_1}$ is the no. of moles of Monomer A and ${n_2}$ is the no. of moles of Dimer B. Hence pressure can be given as, $P = \dfrac{{({n_1} + {n_2})RT}}{V} = \dfrac{{(1 + 0.5) \times 0.0821 \times 546}}{{33.61}} = 2atm$