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
515.1k+ views
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.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

