3.0 mol of gas at a temperature of -120$°$C fills a 2.0 L

container. What is the gas pressure?

The ideal gas law describes the relationship of four state variables for an ideal gas. The four state variables of an ideal gas are the volume V of the container, the pressure p the gas is put on the wall of the container, temperature T of the gas and the number of moles n of the gas in the container.

According to ideal gas law,

$pV=nRT$$\dots \dots \dots \dots \dots \dots \dots \dots \dots \left(1\right)$

Where R is the universal gas constant and in SI unit, its value is

$R=8.31J/mol.K$

Solve equation (1) for p to be in the form

$p=\frac{nRT}{V}$$\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(2\right)$

The temperature has to be in Kelvin. So, it has to be changed from celsius. The units of Kelvin scale is kelvin and symbolised as K. The relation between Kelvin and Celsius is given in the below equation,

$T_K=T_C+273$$\dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \dots \left(3\right)$

By putting the value $T_C=-120°C$into theequation (3) to get $T_K$,

$T_K=T_C+273=-120°C+273=153K$

Convert the unit of V from L to $m^3$by

$V=\left(2L\right)\left(\frac{1m^3}{1000L}\right)=2\times {10}^{-3}{m}^3$

Now, putting the values of n, R, T &V to get the value of p

localid="1648471322667" $\begin{array}{rcl}p& =& \frac{nRT}{V}\\ & =& \frac{\left(30mol\right)\left(8.314J/mol.K\right)\left(153K\right)}{2\times {10}^{-3}{m}^3}\\ & =& 19.08\times {10}^{5}Pa\end{array}$