Question

If the density of a certain gas at ${{30}^{\circ }}C$ and 768 atm pressure is $1.35kg/{{m}^{3}}$, find its density at STP.

Answer 1

As we know that ideal gas equation is also called the general gas equation, which is written as: PV = nRT. To solve this question, we will use the formula of density that is:
$d=\dfrac{PM}{RT}$, where P is the pressure, T is temperature, M is the mass and d is the density.

Complete Step by step solution:
- We are being provided with the values of:
Density (d) = $1.35kg/{{m}^{3}}$
Pressure (p) = 768 torr
Temperature (T) = ${{30}^{\circ }}C$= 30+273 K = 303 K
R =$62.363\times {{10}^{-3}}$
We have to find the density at STP (standard pressure temperature).
- From the ideal gas law, we know that:
PV = nRT
We can write this equation as:
$\begin{aligned} & PV=\dfrac{W}{M}RT \\\ & PM=\dfrac{W}{V}RT \\\ \end{aligned}$Where, $n=\dfrac{W}{M}$
n is the number of moles which is equal to the weight divided by the molecular weight of an atom.
- As we know that density is equal to mass divided by volume that is M/V. So, we can write the equation as:
PM= dRT
$M=\dfrac{dRT}{P}$
- Now, by putting all the values given in the above equation we get:
$\begin{aligned} & M=\dfrac{1.35\times 62.36\times {{10}^{-3}}\times 303}{768} \\\ &\implies 0.033\text{ }kg/mol \\\ \end{aligned}$ - At STP, pressure of 1 bar’
Temperature = 273 K
Pressure = 1 bar
R = $8.314\times {{10}^{-5}}$
M = 0.033 kg/mol
- We will find the value of density as:
PM = dRT
$\begin{aligned} & d=\dfrac{PM}{RT} \\\ & \implies\dfrac{1\times 0.033\times {{10}^{5}}}{8.314\times 27.3} \\\ & \implies1.48\text{ }kg/{{m}^{3}} \\\ \end{aligned}$

- Hence, we can conclude that density at STP will be $=1.48\text{ }kg/{{m}^{3}}$.

Note: - We must convert the temperature given in degrees Celsius into Kelvin.
- We should not forget to write units after solving any question.