Solveeit Logo
Login

Question

Question: The increase in volume of air, when temperature of \[{\rm{600}}\,{\rm{mL}}\] of it, is increased fro...

The increase in volume of air, when temperature of 600 mL{\rm{600}}\,{\rm{mL}} of it, is increased from 27  0C27{\;^{\rm{0}}}{\rm{C}} to 47  0C47{\;^{\rm{0}}}{\rm{C}} under constant pressure is:
(A) 20 mL{\rm{20}}\,{\rm{mL}}
(B) 80 mL{\rm{80}}\,{\rm{mL}}
(C) 40 mL{\rm{40}}\,{\rm{mL}}
(D) 500 mL{\rm{500}}\,{\rm{mL}}

Explanation

Solution

As we know that when any substance is heated to a certain temperature then the particles of substance at that temperature are also heated and they absorb energy and start to colloid with each other and also with the walls of the container.

Complete step by step solution
According to Charles’ law when the pressure of gas remains constant, the volume of a fixed mass of a gas is directly proportional to the absolute temperature.
Thus, if V1{V_1} is the volume of a gas at T1{T_1}temperature and V2{V_2}is the volume of same gas at T2{T_2} temperature at constant pressure, then according to Charles’ law.

V1T1=V2T2 V2=V1×T2T1\dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}\\\ {V_2} = \dfrac{{{V_1} \times {T_2}}}{{{T_1}}}

In the question we are given as
V1=600 mL{V_1} = 60{\rm{0}}\,{\rm{mL}} and T1=27  0C=27+273.15=300.15  K{T_1} = 27{\;^{\rm{0}}}{\rm{C}} = 27 + 273.15 = 300.15\;{\rm{K}}, T2=47  0C=47+273.15=320.15  K{T_2} = 47{\;^{\rm{0}}}{\rm{C}} = 47 + 273.15 = 320.15\;{\rm{K}}
By Putting the above values in equation (i)(i)we get,

V2=600 mL×320.15  K300.15  K =639.98 mL{V_2} = \dfrac{{600\,{\rm{mL}} \times 320.1{\rm{5}}\;{\rm{K}}}}{{300.15\;{\rm{K}}}}\\\ = 639.98\,{\rm{mL}}

So, the change in volume is=639.98 mL−600 mL=39.98 mL≅40 mL = 639.98\,{\rm{mL}} - 600\,{\rm{mL}} = 39.98\,{\rm{mL}} \cong 40\,{\rm{mL}}

Therefore, the correct option is option (C).

Additional information:
Where the absolute temperature is also known as Kelvin scale of temperature. When a graph is plotted at constant pressure between volume (along yyaxis) and temperature (along xxaxis) a straight line when this line is extra-plotted to the lower temperature which represents zero volume. The temperature at which volume becomes zero has been found to be −273.15  0C - 273.15{\;^{\rm{0}}}{\rm{C}}.it is the lowest possible temperature and is known as absolute temperature.
Therefore, we measure as
0  0C=273.15 K0{\;^{\rm{0}}}{\rm{C}} = 273.15\,{\rm{K}}

Note:
The value −273.15  0C - 273.15{\;^{\rm{0}}}{\rm{C}} does not depend upon the nature of gas or the pressure at which the experiment is performed.