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At what temperature the volume of V of a certain mass of gas at ${{27}^{o}}\text{C}$ will be doubled, keeping the pressure constant?
A. ${{54}^{o}}\text{C}$
B. ${{327}^{o}}\text{C}$
C. ${{427}^{o}}\text{C}$
D. ${{527}^{o}}\text{C}$

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Last updated date: 18th Jun 2024
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Answer
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Hint: This question can be solved by simply applying the ideal gas equation. You know that; the gas law which contains pressure, volume and temperature and gives a relation between them is the ideal gas law and the equation is $PV = nRT$.

Complete step by step solution:
The ideal gas equation is shown as:
$PV=nRT$, where
P is the pressure,
V is the volume,
n is the amount of substance,
R is the universal gas constant and
T is the temperature.
As per the question it is given that,
Pressure remains constant.
So, volume will be directly proportional to the temperature.
Also, given that the volume is getting doubled. So, let’s consider the initial volume as ‘V’ and the final volume will be $2\text{V}$.

- Let us consider initial temperature as ‘T’, which is given in the question as ${{27}^{o}}\text{C}$ which can also be represented as $(27\text{+273)K}=300\text{K}$.
Let the final temperature i.e. after doubling the volume be $T'$, which we have to find out.
- From the ideal gas equation shown above:
At T temperature, the equation will be $T=\dfrac{PV}{nR}$ --- (i)
While, after doubling the volume the equation will be:
$T'=\dfrac{P.2V}{nR}=\dfrac{2PV}{nR}$ ---(ii)
After comparing the two equations, we can say that:
$T'=2T$
The value of T is already given.
So, the value of final temperature will be $T'=2\times 300\text{K}=600\text{K}$
In Celsius, the final temperature will be ${{(600-273)}^{o}}\text{C}={{327}^{o}}\text{C}$
So, the correct answer is “Option B”.

Note: It is important to do the unit conversion from Celsius to Kelvin scale and from Kelvin to Celsius scale. To convert Kelvin scale to Celsius scale, $273$ is subtracted while $273$ is added while converting from Celsius to Kelvin scale.