Question

The increase in volume of air, when temperature of $600$mL of it is increased from $27^\circ C$ to$47^\circ C$ under constant pressure , is

Answer 1

Hint: In order to solve this numerical problem we have to apply Charles law which is also known as the law of volumes. Charles law is applied because we need a relation between volume and temperature.

Complete step by step answer:
Step1: Statement of Charles Law:
When the pressure on a sample of dry gas is held constant, the temperature and volume will be in direct proportion.
Step 2: This relationship of direct proportion can be written as:
$V \propto T$
So this means:
$\dfrac{V}{T} = k$ or $V = kT$
Where: V is the volume of the gas,
T is the temperature of the gas (measured in Kelvin)
K is a non-zero constant.
Step3: In the statement we are provided with initial volume ${V_1}$ , initial temperature ${T_1}$ and final temperature ${T_2}$ we have to calculate the value of final volume ${V_2}$
${T_1} = 27^\circ C$
${T_2} = 47^\circ C$
${V_1} = 600\,mL$
For comparing the same substance under two set of conditions, the law can be written as:
$\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$ or $\dfrac{{{V_2}}}{{{V_1}}} = \dfrac{{{T_2}}}{{{T_1}}}$ ………..$(i)$
 Converting temperature from celsius to Kelvin, using formula
 $K = x^\circ C + 273$
$\therefore {T_1} = 27 + 273 = 300K$
${T_2} = 47 + 273 = 320K$
Substituting the values in eq $(i)$
${V_2} = \dfrac{{{T_2}}}{{{T_1}}} \times {V_1}$
$= \dfrac{{320}}{{300}} \times 600$
${V_2} = 600mL$
Increase in volume of air = Final volume – Initial volume
$= {V_2} - {V_1}$
$= 640 - 600$
$= 40mL$

Note: This law describes how a gas expands as the temperature increases; conversely a decrease in temperature will lead to a decrease in volume. Everyday examples of Charle’s law : In winters as the temperature decreases, when we take a basketball outside in the ground the ball shrinks. This is the only reason why to check the pressure in the car tyres when going outside in the cold days.