Question

The volume of a certain gas was found to be $560\;c{m^3}$ when the pressure was $600\;mm$ . If the pressure decreases by $40\%$ , then find the new volume of the gas.

Answer 1

Hint : Boyle’s law: It states that when the temperature and the number of moles of an ideal gas for a system is constant, then the pressure of the gas is inversely proportional to the volume of the gas i.e., if the pressure of the gas increases, then a decrease in the volume of gas will be observed.

Complete Step By Step Answer:
As per question, the given data is as follows:
Initial volume of gas ${V_1} = 560\;c{m^3}$
Initial pressure of gas ${P_1} = 600\;mm$
As the final pressure of the gas is $40\%$ less than the initial pressure. Therefore, it can be written as:
${P_2} = {P_1} - \dfrac{{40}}{{100}} \times {P_1}$
Substituting values:
$\Rightarrow {P_2} = 600 - 0.4 \times 600$
$\Rightarrow {P_2} = 360\;mm$
According to Boyle’s law, the pressure of the gas inversely varies with the volume of the gas. It can be expressed as follows:
$P \propto \dfrac{1}{V}$
On removing the proportionality sign, a proportionality constant is introduced to the expression as follows:
$P = \dfrac{k}{V}$
$\Rightarrow PV = k\;\; - (i)$
At initial condition of gases, the equation (i) can be expressed as ${P_1}{V_1} = k\;\;\; - (ii)$
At final condition of gases, the equation (i) can be expressed as ${P_2}{V_2} = k\;\;\; - (iii)$
When equations (ii) and (iii) are compared, then the relation which is obtained as follows:
${P_1}{V_1} = {P_2}{V_2}$
Substituting values:
$600 \times 560 = 360 \times {V_2}$
$\Rightarrow {V_2} = \dfrac{{600 \times 560}}{{360}}$
$\Rightarrow {V_2} = 933.33\;c{m^3}$
Hence the new volume of the gas is $933.33\;c{m^3}$ .

Note :
It is important to note that Boyle’s law is only applicable to an isothermal process in which the number of moles of gas are fixed. Also, if a graph is plotted between pressure and volume for a fixed mass of gas in an isothermal process, then an exponential curve is obtained as follows: