
During an experiment an ideal gas is found to obey an additional law \[V{P^2} = \] constant. The gas is initially at temperature T and volume V, when it expand to volume $2V,$ the resulting temperature is ${T_2}:$
$
A.{\text{ }}\dfrac{T}{2} \\
B.{\text{ 2T}} \\
{\text{C}}{\text{. }}\sqrt 2 T \\
D.{\text{ }}\dfrac{T}{{\sqrt 2 }} \\
$
Answer
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Hint: Here, we will use the law of ideal gas. That is $PV = nRT$ where P is pressure, V is volume, n is the present number of moles of gas, R is ideal gas constant and T is the temperature. Then by using the given condition and finding the correlation between the two, will find the required change in Temperature.
Complete step by step answer:
Let, the initial volume be $ = V$
The initial temperature be $ = T$
The final volume be $ = 2V$
The final temperature be $ = {T_2}$
Now, by using the law of ideal gas equation –
$PV = nRT$ ….. (a)
$ \Rightarrow P = \dfrac{{nRT}}{V}$ ……. (b)
And given that, \[V{P^2} = \] constant
Re-writing the above equation –
\[VP \times P = \] Constant
\[PV \times P = \] Constant
By placing value from the equation (a)
$nRT \times P = $ Constant
Place the value of “P” from the equation (b)
$nRT \times \dfrac{{nRT}}{V} = $ Constant
$\dfrac{{{{(nRT)}^2}}}{V} = $ Constant
Since, n and R are constant
$ \Rightarrow \dfrac{{{T^2}}}{V} = $ Constant
Now, when volume extends from V to $2V$ and temperature to ${T_2}$
$ \Rightarrow \dfrac{{{T_2}^2}}{{2V}} = $Constant
$ \Rightarrow \dfrac{{{T^2}}}{V} = $$\dfrac{{{T_2}^2}}{{2V}}$
Do cross multiplication and make unknown temperature ${T_2}$ the subject
${T_2} = \sqrt {\dfrac{{2V \times {T^2}}}{V}} $
Square and square-root cancel each other on both the sides of the equation.
$ \Rightarrow {T_2} = \sqrt 2 T$.
Hence, from the given multiple choices – the option C is the correct answer.
Note:
To solve these types of word problems always remember the basic five types of gas laws which deal with how gases behave with respect to the pressure, volume and temperature.
- Boyle’s law
- Charles’ law
- Gay-Lussac’s law
- Combined law
- Ideal Gas law
Complete step by step answer:
Let, the initial volume be $ = V$
The initial temperature be $ = T$
The final volume be $ = 2V$
The final temperature be $ = {T_2}$
Now, by using the law of ideal gas equation –
$PV = nRT$ ….. (a)
$ \Rightarrow P = \dfrac{{nRT}}{V}$ ……. (b)
And given that, \[V{P^2} = \] constant
Re-writing the above equation –
\[VP \times P = \] Constant
\[PV \times P = \] Constant
By placing value from the equation (a)
$nRT \times P = $ Constant
Place the value of “P” from the equation (b)
$nRT \times \dfrac{{nRT}}{V} = $ Constant
$\dfrac{{{{(nRT)}^2}}}{V} = $ Constant
Since, n and R are constant
$ \Rightarrow \dfrac{{{T^2}}}{V} = $ Constant
Now, when volume extends from V to $2V$ and temperature to ${T_2}$
$ \Rightarrow \dfrac{{{T_2}^2}}{{2V}} = $Constant
$ \Rightarrow \dfrac{{{T^2}}}{V} = $$\dfrac{{{T_2}^2}}{{2V}}$
Do cross multiplication and make unknown temperature ${T_2}$ the subject
${T_2} = \sqrt {\dfrac{{2V \times {T^2}}}{V}} $
Square and square-root cancel each other on both the sides of the equation.
$ \Rightarrow {T_2} = \sqrt 2 T$.
Hence, from the given multiple choices – the option C is the correct answer.
Note:
To solve these types of word problems always remember the basic five types of gas laws which deal with how gases behave with respect to the pressure, volume and temperature.
- Boyle’s law
- Charles’ law
- Gay-Lussac’s law
- Combined law
- Ideal Gas law
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