Question

To what temperature must a neon gas sample be heated to double the pressure, if the initial volume of a gas at ${75^0}{\text{C}}$ is decreased by $15\%$ ?
A.319 $^{\text{o}}{\text{C}}$
B.592 $^{\text{o}}{\text{C}}$
C.128 $^{\text{o}}{\text{C}}$
D.60 $^{\text{o}}{\text{C}}$
E.90 $^{\text{o}}{\text{C}}$

Answer 1

To solve this question, we require knowledge of the combined gas law which is a result of the combination of Charles' Law and Boyle's Law.
Formula used: $\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$
Where, ${{\text{V}}_1}$ is the initial volume of the gas, ${{\text{P}}_1}{\text{ and }}{{\text{T}}_1}$ are the initial pressure and temperature, ${{\text{V}}_2}{\text{ , }}{{\text{P}}_2}{\text{ and }}{{\text{T}}_2}$ are the final volume, pressure and temperature respectively.

Complete step by step answer:
Let the initial volume, ${{\text{V}}_1}$ of the gas be V.
Initial pressure, ${{\text{P}}_1}$ be P and the initial temperature,
${{\text{T}}_1}$= ${75^0}{\text{C}}$=$273 + 75 = 348{\text{K}}$
Therefore the final volume, according to the question,
${{\text{V}}_{\text{2}}}$=${\text{V}} - \dfrac{{15}}{{100}}{\text{V = }}\dfrac{{17}}{{20}}{\text{V}}$
And pressure ${{\text{P}}_2}$ = 2P
Therefore putting the values of the parameters in the combined gas equation, we get,
$\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$
$\Rightarrow \dfrac{{{\text{PV}}}}{{348}}{\text{ = }}\dfrac{{{\text{2P}} \times {\text{17V}}}}{{{\text{20T}}}}$
Rearranging this, we get:
$\Rightarrow {{\text{T}}_2} = \dfrac{{348 \times 2 \times 17}}{{20}}$
$\Rightarrow {{\text{T}}_2} = 591.6{\text{K}}$
Hence, the temperature of the gas after expansion is equal to $591.6{\text{K}} \sim 592{\text{K}}$
$\Rightarrow 591.6-273$
$\Rightarrow 319^0C$

Hence, the correct option is option A.

Note:
Charles’ Law states that, “When the pressure of a fixed mass of a dry gas is kept constant, then the change in the volume is directly proportional to the temperature of the gas in the Kelvin scale.”
Mathematically,
${\text{V}} \propto {\text{T}} \Rightarrow \dfrac{{{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$
Boyle’s Law states that, “Under constant temperature, the pressure of a dry gas is inversely proportional to the volume of the gas”.
Mathematically,
${\text{P}} \propto \dfrac{{\text{1}}}{{\text{V}}} \Rightarrow {{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}{\text{ = }}{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}$
Combining both these laws, we get the combined gas law as stated above,
$\dfrac{{{{\text{P}}_{\text{1}}}{{\text{V}}_{\text{1}}}}}{{{{\text{T}}_{\text{1}}}}}{\text{ = }}\dfrac{{{{\text{P}}_{\text{2}}}{{\text{V}}_{\text{2}}}}}{{{{\text{T}}_{\text{2}}}}}$