Question

At a certain temperature for which $RT = 25lit.atm.mo{l^{ - 1}}$, the density of gas, in $gmli{t^{ - 1}}$, is $d = 2.00P + 0.020{P^2}$, where P is the pressure in atmosphere. The molecular weight of the gas in gm $mo{l^{ - 1}}$ is:

Answer 1

Hint:The law which is related to the ideal gases is known as ideal gas law. The ideal gas equation gives a relation between the pressure, the volume of gas with the universal gas constant, and temperature. For the ideal gas equation, in the linear dependence with positive slope, at the Y-intercept $P/T \to 0$ and $p \to 0$ all the gases in the system show ideal behavior.

Complete step by step answer:
Given,
RT = 25lit.atm.$mo{l^{ - 1}}$
$d = 2.00P + 0.020{P^2}$
The relationship between the molecular weight of the compound and density is given by as shown below.
$D = \dfrac{{MP}}{{RT}}$
Where,
D is the density
M is the molecular mass
P is the pressure
R is the universal gas constant
T is the temperature
This equation helps to know the relation between pressure, temperature, and density of the gas. This equation does not depend on the volume of gas. It helps to determine the value of density for the gas when pressure and temperature are given. The molar mass can be determined when density is given.
${\mathop {Lt}\limits_{P \to O} ^{\dfrac{d}{P}}} = \dfrac{M}{{RT}}$
To calculate the molecular weight substitute the values in the equation
$M = 2RT$
$\Rightarrow M = 2 \times 25$
$\Rightarrow M = 50$
The molecular weight of the gas is 50 g/mol.

Note:
The molecular weight or molar mass of any gas is defined as the mass carried by one particle of the gas multiplied with the value of Avogadro’s number that is $6.02 \times {10^{23}}$. The average molar mass for the gases present in the mixture is given as the sum of the mole fraction multiplied with their molar masses.