Question

At ${0^ \circ }{\text{C}}$ and $1{\text{atm}}$ pressure, a gas occupies $100{\text{cc}}$. If the pressure is increased to one and half time and temperature is increased by one-third of absolute temperature, then final volume of the gas will be:
A. $80{\text{cc}}$
B. $88.9{\text{cc}}$
C. $66.7{\text{cc}}$
D. $100{\text{cc}}$

Answer 1

Ideal gas equation is the principle behind this question. Ideal gas is a perfect which obeys Boyle’s law, Charles’ law and Avogadro’s law. This is modelled on kinetic theory of gases. When temperature, pressure and volume is changed, there is no change in the number of moles of gas.

Complete step by step answer:
Given data:
Initial temperature, ${{\text{T}}_1} = {0^ \circ }{\text{C = 273}}{\text{.15K}}$
Initial pressure, ${{\text{P}}_1} = 1{\text{atm}}$
Initial volume, ${{\text{V}}_1} = 100{\text{cc}}$
Ideal gas equation is derived from three main laws-Boyle’s ;aw, Charles’ law and Avogadro’s law. Boyle’s law gives the relationship between pressure and volume. Charles’ law gives the relationship between volume and temperature. Avogadro’s law gives the relationship between the volume and number of moles of gas.
It is given that initial temperature, ${{\text{T}}_1} = {0^ \circ }{\text{C = 273}}{\text{.15K}}$
Initial pressure, ${{\text{P}}_1} = 1{\text{atm}}$
Initial volume, ${{\text{V}}_1} = 100{\text{cc}}$
Based on ideal gas law,
${{\text{P}}_1}{{\text{V}}_1} = {\text{nR}}{{\text{T}}_1}$, where ${\text{n}}$is the number of moles of gas and ${\text{R}}$is the gas constant.
${\text{n}}$and ${\text{R}}$ are the same in the given conditions.
Therefore, ${\text{nR}} = \dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{T}}_1}}} \to \left( 1 \right)$
Now the temperature, pressure and volume has changed.
Final temperature is increased by one-third of absolute temperature.
Therefore, final temperature can be expressed as ${{\text{T}}_2} = \left( {1 + \dfrac{1}{3}} \right) \times 273.15{\text{K = }}\dfrac{4}{3} \times 273.15{\text{K = 364}}{\text{.2K}}$
Initial pressure increased to one and half time to get final pressure.
i.e. Final pressure, ${{\text{P}}_2} = \left( {1 + \dfrac{1}{2}} \right) \times 1{\text{atm = }}\dfrac{3}{2}{\text{atm = 1}}{\text{.5atm}}$
${{\text{P}}_2}{{\text{V}}_2} = {\text{nR}}{{\text{T}}_2}$
${\text{nR}} = \dfrac{{{{\text{P}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}} \to \left( 2 \right)$
Equating $\left( 1 \right)$ and $\left( 2 \right)$, we get
$\dfrac{{{{\text{P}}_1}{{\text{V}}_1}}}{{{{\text{T}}_1}}} = \dfrac{{{{\text{P}}_2}{{\text{V}}_2}}}{{{{\text{T}}_2}}}$
Substituting the values in the above equation, we get
$\dfrac{{1{\text{atm}} \times 100{\text{cc}}}}{{273.15{\text{K}}}} = \dfrac{{1.5{\text{atm}} \times {{\text{V}}_2}}}{{364.2{\text{K}}}}$
On further simplifying, we get
$0.366 = \dfrac{{1.5{{\text{V}}_2}}}{{364.2}}$
Thus final volume can be calculated.
Final volume, ${{\text{V}}_2} = \dfrac{{0.366 \times 364.2}}{{1.5}} = 88.86{\text{cc}}$
Hence the final volume of the gas, ${{\text{V}}_2} = 88.86{\text{cc}}$

Hence, Option B is the correct option.

Additional information:
Ideal gases do not condense into a liquid at low temperatures. They do not have an attractive or repulsive force between the particles. It is composed of particles that have no volume.

Note:
This can also be solved just by using Boyle’s law and Charles’ law since it explains about the relationship between the volume, temperature and pressure. No gas perfectly follows all the gas laws under all conditions. We assume that they follow the gas laws to study the behavior of gases.