Question

Find the temperature at which volume and pressure of $28g$ nitrogen gas becomes $10d{m^3}$ and $2.46atm$ respectively.
A.$1200K$
B.$600K$
C.$300K$
D.$900K$

Answer 1

Hint: The relation between the temperature, pressure and volume of a gas is given by the formula,
$PV = nRT$
This equation is the ideal gas equation.
The units of pressure, volume and temperature considered in the above equation are $atm$, $L$ and $K$ respectively. So, the corresponding unit of R is $Latm{K^{ - 1}}mo{l^{ - 1}}$.

Complete step by step answer:
Nitrogen gas exists as ${N_2}$. So, the molar mass of nitrogen gas is
$ = 2 \times 14g$
$ \Rightarrow 28g$
In the question, $28g$ nitrogen gas is present, therefore the number of moles of gas (n) present can be calculated by dividing the given mass of the gas by the molar mass of the gas.
So, $n = \dfrac{{28}}{{28}}$
$ \Rightarrow n = 1$ mole
Volume of the gas is given to be $10d{m^3}$. We will convert $d{m^3}$ into litre.
As, $1d{m^3} = 1L$
So, $10d{m^3} = 10L$
Pressure of the gas is $2.46atm$ and R is the universal gas constant whose value is $0.0821Latm{K^{ - 1}}mo{l^{ - 1}}$.
Putting these values in the ideal gas equation,
$PV = nRT$
$ \Rightarrow 2.46 \times 10 = 1 \times 0.0821 \times T$
$ \Rightarrow T = \dfrac{{2.46 \times 10}}{{0.0821}}K$
$ \Rightarrow T = 300K$

Thus, the answer is option C.

Note:
While considering the molar mass of the gas we should keep in mind the atomic state in which the gas exists, because gases like nitrogen, oxygen, etc. exist as diatomic gases (${N_2},{O_2}$); therefore, their diatomic masses should be considered while solving the question.
Also, the value of R is taken to be $0.0821Latm{K^{ - 1}}mo{l^{ - 1}}$ when the units of P and V are $atm$ and $L$ respectively. Whereas the value of R as $8.314J{K^{ - 1}}mo{l^{ - 1}}$ is taken when the units of P and V are $Pa$ and ${m^3}$ respectively.