
For the specific heat of 1 mole of an ideal gas at constant pressure $\left( {{{\text{C}}_{\text{P}}}} \right)$ and at constant volume $\left( {{{\text{C}}_{\text{V}}}} \right)$ , which one of the given options is correct?
A. ${{\text{C}}_{\text{P}}}$ of hydrogen gas is $\dfrac{5}{2}R$ .
B. ${{\text{C}}_{\text{V}}}$ of hydrogen gas is $\dfrac{7}{2}R$ .
C. ${H_2}$ has very small value of ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ .
D. ${{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = 1.99{\text{ cal/mol K}}$ for ${H_2}$ .
Answer
161.1k+ views
Hint:To solve the given question, start checking each of the given options. Find out the value of ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ of an ideal diatomic gas. Also remember that ${{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R$ , where $R = 8.314{\text{ Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
Complete answer:
We’ll check each option to determine the correct one.
Hydrogen gas is a diatomic gas. ${{\text{C}}_{\text{P}}}$ of an ideal diatomic gas is $\dfrac{7}{2}R$ , where $R$ is the universal gas constant.
Hence, the first option is incorrect.
${{\text{C}}_{\text{V}}}$ of an ideal diatomic gas is $\dfrac{5}{2}R$ , where $R$ is the universal gas constant.
Hence, the second option is incorrect as well.
Clearly, the third option is incorrect as well.
Now, for one mole of an ideal gas, we know that ${{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R = 8.314{\text{ Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
There are $4.186$ joules in one calorie, hence, converting $R$ in ${\text{cal mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ ,
${{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = \dfrac{{8.314}}{{4.186}} = 1.99{\text{ cal mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$
Thus, the correct option is D.
Note: To solve the given question, just remember the relation between ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ which is given by \[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R\] . Note that the value of $R$ provided in the options using this formula is in units of calories while using the relation, we obtain it in units of joules. Hence, perform basic maths and convert the relation in calories per mole per kelvin to get the required answer.
Complete answer:
We’ll check each option to determine the correct one.
Hydrogen gas is a diatomic gas. ${{\text{C}}_{\text{P}}}$ of an ideal diatomic gas is $\dfrac{7}{2}R$ , where $R$ is the universal gas constant.
Hence, the first option is incorrect.
${{\text{C}}_{\text{V}}}$ of an ideal diatomic gas is $\dfrac{5}{2}R$ , where $R$ is the universal gas constant.
Hence, the second option is incorrect as well.
Clearly, the third option is incorrect as well.
Now, for one mole of an ideal gas, we know that ${{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R = 8.314{\text{ Jmo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ .
There are $4.186$ joules in one calorie, hence, converting $R$ in ${\text{cal mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$ ,
${{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = \dfrac{{8.314}}{{4.186}} = 1.99{\text{ cal mo}}{{\text{l}}^{ - 1}}{{\text{K}}^{ - 1}}$
Thus, the correct option is D.
Note: To solve the given question, just remember the relation between ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ which is given by \[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R\] . Note that the value of $R$ provided in the options using this formula is in units of calories while using the relation, we obtain it in units of joules. Hence, perform basic maths and convert the relation in calories per mole per kelvin to get the required answer.
Recently Updated Pages
A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

A wave is travelling along a string At an instant the class 11 physics JEE_Main

The length of a conductor is halved its conductivity class 11 physics JEE_Main

Two billiard balls of the same size and mass are in class 11 physics JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Uniform Acceleration

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

Motion in a Straight Line Class 11 Notes: CBSE Physics Chapter 2

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line
