
One mole of an ideal gas undergoes a process $P = \dfrac{{{P_0}}}{{1 + {{(\dfrac{V}{{{V_0}}})}^2}}},$ where ${P_0},{V_0}$ are constants. Find the temperature of the gas when $V = {V_0}$
A. $\dfrac{{{P_0}{V_0}}}{R}$
B. $\dfrac{{2{P_0}{V_0}}}{{3R}}$
C. $\dfrac{{2{P_0}{V_0}}}{R}$
D. $\dfrac{{{P_0}{V_0}}}{{2R}}$
Answer
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Hint: In order to solve this question, we should know that, an ideal gas is one which obeys ideal gas equation and one mole of a gas simple means the number of moles of gas at given pressure, volume, temperature is one. Here, we will use the general ideal gas equation to find the temperature of the ideal gas under given parametric conditions.
Formula Used:
An ideal gas obeys the ideal gas equation as,
$PV = nRT$
where,$P$ is the pressure of the gas, $V$ is the volume of the gas, $T$ is the temperature at which Pressure and Volume are measured of the gas, $n$ is the number of moles of the gas and $R$ is known as the Universal Gas constant.
Complete step by step answer:
According to the question we have given that, relation between pressure and volume as $P = \dfrac{{{P_0}}}{{1 + {{(\dfrac{V}{{{V_0}}})}^2}}}$ we need to find temperature at condition $V = {V_0}$ so, put this value we get,
$ \Rightarrow P = \dfrac{{{P_0}}}{{1 + {{(\dfrac{V}{{{V_0}}})}^2}}}$ at $V = {V_0}$
$ \Rightarrow P = \dfrac{{{P_0}}}{{1 + {{(\dfrac{{{V_0}}}{{{V_0}}})}^2}}}$
$ \Rightarrow P = \dfrac{{{P_0}}}{{1 + 1}}$
$ \Rightarrow P = \dfrac{{{P_0}}}{2}$
Now, we have pressure of the gas is $\dfrac{{{P_0}}}{2}$ volume of the gas is ${V_0}$ and number of mole is given as $n = 1\,mole$ and T be the temperature so using, $PV = nRT$ we have,
$\dfrac{{{P_0}}}{2}{V_0} = 1RT$
$ \therefore T = \dfrac{{{P_0}{V_0}}}{{2R}}$
Hence, the correct option is D.
Note: It should be remembered that, there are some gases which do not obey ideal gas equation and are known as Real gases as well as Not any gas is ideally perfect and Universal gas constant has a fixed value of $R = 8.314\,Jmo{l^{ - 1}}{K^{ - 1}}$ and ideal gas equation is the basis of foundation of further equations of thermodynamics.
Formula Used:
An ideal gas obeys the ideal gas equation as,
$PV = nRT$
where,$P$ is the pressure of the gas, $V$ is the volume of the gas, $T$ is the temperature at which Pressure and Volume are measured of the gas, $n$ is the number of moles of the gas and $R$ is known as the Universal Gas constant.
Complete step by step answer:
According to the question we have given that, relation between pressure and volume as $P = \dfrac{{{P_0}}}{{1 + {{(\dfrac{V}{{{V_0}}})}^2}}}$ we need to find temperature at condition $V = {V_0}$ so, put this value we get,
$ \Rightarrow P = \dfrac{{{P_0}}}{{1 + {{(\dfrac{V}{{{V_0}}})}^2}}}$ at $V = {V_0}$
$ \Rightarrow P = \dfrac{{{P_0}}}{{1 + {{(\dfrac{{{V_0}}}{{{V_0}}})}^2}}}$
$ \Rightarrow P = \dfrac{{{P_0}}}{{1 + 1}}$
$ \Rightarrow P = \dfrac{{{P_0}}}{2}$
Now, we have pressure of the gas is $\dfrac{{{P_0}}}{2}$ volume of the gas is ${V_0}$ and number of mole is given as $n = 1\,mole$ and T be the temperature so using, $PV = nRT$ we have,
$\dfrac{{{P_0}}}{2}{V_0} = 1RT$
$ \therefore T = \dfrac{{{P_0}{V_0}}}{{2R}}$
Hence, the correct option is D.
Note: It should be remembered that, there are some gases which do not obey ideal gas equation and are known as Real gases as well as Not any gas is ideally perfect and Universal gas constant has a fixed value of $R = 8.314\,Jmo{l^{ - 1}}{K^{ - 1}}$ and ideal gas equation is the basis of foundation of further equations of thermodynamics.
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