Question

The molecular weight of a gas is $40$. At $400\text{K}$, if $120g$ of this gas has a volume of $20\text{ litres}$, the pressure of the gas is?
A. $5.02\text{ atm}$
B. $\text{0}\text{.546 atm}$
C. $\text{4}\text{.92 atm}$
D. $\text{49}\text{.6 atm}$

Answer 1

Hint: To solve this question, simply apply the ideal gas equation. The ideal gas equation is given by $PV=nRT$. Keep in mind that; one mole of gas is the molar mass of a gas. Use the standard temperature and pressure values to find the required answer.

Complete step by step solution:
First of all; let us have some knowledge about the ideal gas equation. So, the ideal gas equation has pressure, temperature, moles, and volume. The equation for the ideal gas is given by:
$PV=nRT$, where
P is the pressure,
V is the volume,
n is the number of moles (i.e. the amount of substance),
R is the universal gas constant (whose value is $0.0821\text{ atm}\text{.L}\text{.mo}{{\text{l}}^{-1}}\text{.}{{\text{K}}^{-1}}$) and
T is the temperature.
Given that,
The molecular weight of a gas is $40$and at $400K$, the weight of the gas given is $120g$ and volume is $20$ liters. The pressure (consider as P) of the gas is to be calculated.
We know that; one mole of gas means the molar mass of the gas and equals to the ratio of weight given to that of the molecular mass of that substance.
So, here the mole of gas will be $=\dfrac{\text{mass given}}{\text{molecular weight of the gas}}=\dfrac{120}{40}=3$
Now by applying the ideal gas equation, we will get the pressure value as:
$PV=nRT$
Then, $P=\dfrac{nRT}{V}$
By putting the values,
$P=\dfrac{3\times 0.0821\times 400}{20}=4.926\text{ atm}\cong \text{4}\text{.92 atm}$
The value of pressure is $4.926\text{ atm}$.

Hence, the correct option is C.

Note: In ideal gas related questions, the formula that is to be kept in mind is $PV=nRT$. From this single formula, other formulas can be derived. The universal gas constant that is denoted as R has different values like $0.0821\text{ atm}\text{.L}\text{.mo}{{\text{l}}^{-1}}.{{\text{K}}^{-1}}$, $8.314\text{ J}\text{.mo}{{\text{l}}^{-1}}\text{.}{{\text{K}}^{-1}}$ and $\text{1}\text{.987 cal}\text{.mo}{{\text{l}}^{-1}}\text{.}{{\text{K}}^{-1}}$. The units have to be considered.