Question

Given, $35^{\circ}C$ and $1$ atmosphere pressure occupies a volume of ${\text{3}}{\text{.75 L}}$. At what temperature should the gas be kept, if it is required to reduce the volume a ${\text{3}}{\text{.0L}}$ at the same pressure?
A) $-26.6^{\circ}C$
B) $0^{\circ}C$
C) $.98^{\circ}C$
D) $28^{\circ}C$

Answer 1

Charles is one of the important laws in the gaseous state. Charles' law was proposed by J.A.C. Charles. This law is used to study the relationship volume of a gas and its temperature. In Charles' law the mass of the system and pressure of the system is constant. The Combination of Charles law, Boyle's law and Avogadro’s hypothesis is known as the ideal gas equation.
Formula used:
Charles law is nothing but once volume of gas increases, temperature of gas also increases and once volume of gas decreases, temperature of gas also decreases. Finely, volume of gas is directly proportional to the temperature of the gas at constant pressure and constant mass of the system.
Charles law
${\text{V}} \propto {\text{T}}$ at constant pressure
$\dfrac{{\text{V}}}{{\text{T}}}{\text{ = constant}}$
Here, V is volume of gas
The temperature of the gas is ${\text{T}}$.
Formula for convert degree Celsius to kelvin in temperature
${\text{kelvin = degree + 273}}$
Formula for convert kelvin Celsius to degree in temperature
${\text{degree = kelvin - 273}}$

Complete step by step answer:
Given,
Initial temperature of gas is ${\text{35}}^\circ {\text{C}}$
Formula for convert degree Celsius to kelvin in temperature
${\text{kelvin = degree + 273}}$
${\text{ = 35 + 273}}$
${\text{ = 308k}}$
Final temperature of gas is $X$
Initial volume of gas is ${{\text{V}}_1}{\text{ = 3}}{\text{.75 L}}$
Final volume of gas is ${{\text{V}}_2}{\text{ = 3}}{\text{.0 L}}$
Charles law
${\text{V}}\propto {\text{T at constant pressure}}$
$\dfrac{{\text{V}}}{{\text{T}}}{\text{ = constant}}$
$\dfrac{{{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$
$\dfrac{{{\text{3}}{\text{.75}}}}{{308}}{\text{ = }}\dfrac{{{\text{3}}{\text{.0}}}}{{{{\text{T}}_{\text{2}}}}}$
${{\text{T}}_{\text{2}}}{\text{ = }}\dfrac{{{\text{3}}{\text{.0}} \times 308}}{{{\text{3}}{\text{.75}}}}$
${{\text{T}}_{\text{2}}}{\text{ = }}\dfrac{{924}}{{{\text{3}}{\text{.75}}}}$
${{\text{T}}_{\text{2}}}{\text{ = 246}}{\text{.4k}}$
Temperature at degree,
${\text{degree = kelvin - 273}}$
${{\text{T}}_{\text{2}}}{\text{ = 246}}{\text{.4 - 273}}$
${{\text{T}}_{\text{2}}}{\text{ = - 26}}{\text{.6}}^\circ {\text{C}}$
A sample of gas at $35^{\circ}C$and $1$ atmosphere pressure occupies a volume of ${\text{3}}{\text{.75 L}}$. If it is required to reduce the volume a ${\text{3}}{\text{.0L}}$ at the same pressure, temperature should the gas be $-26.6^{\circ}C$

Hence, option A is correct.

Note:
We have to know that the ideal gas equation depends on the pressure, temperature, number of moles, volume of the gas molecules in ideal condition. The Combination of Charles law, Boyle's law and Avogadro’s hypothesis is known as the ideal gas equation. The ideal gas equation is ${\text{PV = nRT}}$. Here, the pressure of the gas is P. The volume of the gas is V. The temperature of the gas in kelvin is T. Gas constant is R. The number of moles of the Gas molecules is n.