Question

When the pressure of 5 L of ${N_2}$ is doubled and its temperature is raised from 300 K to 600 K, the volume of gas would be,
A. 10 L
B. 5 L
C. 15 L
D. 20 L

Answer 1

We know that, when pressure applied on the gas increases volume of the gas decreases and when pressure decreases volume of gas increases. Similarly, when temperature increases volume of gas increases and when pressure decreases volume of the gas decreases we know temperature, pressure and volume of gas are dependent to each other and we establish a relation between them given by,
$PV = nRT$
Where, $P =$ Pressure exerted by the gas
$V =$ Volume of the gas
$n =$ Moles of gas
$R =$ Universal gas constant
$T =$Temperature of the gas
This equation is known as the ideal gas equation.

Complete step-by-step answer: According to Boyle's law : "At constant temperature, the volume of a sample of a gas varies inversely with the pressure".
$V \propto \dfrac{1}{P}$
When temperature and number of moles kept constant

According to Charles' law : It relates the temperature and volume of a given mass of a gas at constant pressure
"At constant pressure, the volume of a sample of a gas varies proportionally with the temperature".
$V \propto T$
On combining Boyle's law and Charles law we get,
$V \propto \dfrac{T}{P}$
Or,
$PV \propto T$
$PV = nRT$
This equation is known as the ideal gas equation.
In the question, we are given
Initial volume of gas, ${V_i} = 5{\text{ L}}$
Initial temperature of gas, ${T_i} = 600{\text{ K}}$
Final temperature of gas, ${T_f} = 600{\text{ K}}$
$\dfrac{{{P_f}}}{{{P_i}}} = 2$
From ideal gas equation
$PV = nRT$
${P_i}{V_i} = nR{T_i}$
${P_i} \times 5 = nR \times 300$
$\dfrac{{{P_i}}}{{60}} = nR{\text{ }}...\left( 1 \right)$
For second condition,
Using the ideal gas equation
$PV = nRT$
${P_f}{V_f} = nR{T_f}$
${P_f}{V_f} = nR \times 600$
$\dfrac{{{P_f}{V_f}}}{{600}} = nR{\text{ }}...\left( 2 \right)$
On dividing equation (2) by equation (1) we get,
$\dfrac{{{P_f}{V_f}}}{{{P_i} \times 10}} = 1$
${V_f} = 10 \times \dfrac{{{P_i}}}{{{P_f}}}$
By given data,
${V_f} = 10 \times \dfrac{1}{2}$
${V_f} = 5{\text{ L}}$
So, the volume of gas at 600 K is 5 L

Hence, option (B) is correct.

Note: Ideal gas equation can be applied on only ideal gases, a normal gas acts as an ideal gas at high temperature and low pressure. Gas whose molecules occupy negligible space and have no interaction between molecules of it is known as an ideal gas.