
The value of the gas constant $\left( R \right)$ calculated from the perfect gas equation is $8.32{\text{ Joule/gm mol K}}$ , whereas its value calculated from the knowledge of ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ of the gas is ${\text{1}}{\text{.98 cal/gm mol K}}$ . What is the value of $J$ from this data?
A. $4.16{\text{ J/cal}}$
B. $4.18{\text{ J/cal}}$
C. $4.20{\text{ J/cal}}$
D. $4.22{\text{ J/cal}}$
Answer
145.2k+ views
Hint:${{\text{C}}_{\text{P}}}$ is the molar heat capacity of a gas at constant pressure and ${{\text{C}}_{\text{V}}}$ is the molar heat capacity of the gas at constant volume. For an ideal gas, the relation between ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ is given by \[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R\] .
Formula used:
For an ideal gas, the relation between ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ is given by
\[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R\] .
Complete answer:
For an ideal gas, the relation between ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ is given by:
\[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R\] …(1)
Here, $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$.
However, in the given question, we are provided this value of $R$ in the unit of calories.
Hence, dividing the right-hand side of the relation in equation (1) by 1 Joule to get it in the form of calories,
\[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = \dfrac{R}{J}\]
Now, the given value of \[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}}\] is ${\text{1}}{\text{.98 cal/gm mol K}}$ .
Thus, substituting all the values, we get:
${\text{1}}{\text{.98 cal/gm mol K}} = \dfrac{{8.32{\text{ Joule/gm mol K}}}}{J}$
On simplifying further, we get:
$J = \dfrac{{8.32}}{{1.98}} = 4.20{\text{ J/cal}}$
Thus, the correct option is C.
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 question 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 to get the required answer.
Formula used:
For an ideal gas, the relation between ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ is given by
\[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R\] .
Complete answer:
For an ideal gas, the relation between ${{\text{C}}_{\text{P}}}$ and ${{\text{C}}_{\text{V}}}$ is given by:
\[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = R\] …(1)
Here, $R$ is the universal gas constant having a value of $8.314{\text{ J}}{{\text{K}}^{ - 1}}{\text{mo}}{{\text{l}}^{ - 1}}$.
However, in the given question, we are provided this value of $R$ in the unit of calories.
Hence, dividing the right-hand side of the relation in equation (1) by 1 Joule to get it in the form of calories,
\[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}} = \dfrac{R}{J}\]
Now, the given value of \[{{\text{C}}_{\text{P}}} - {{\text{C}}_{\text{V}}}\] is ${\text{1}}{\text{.98 cal/gm mol K}}$ .
Thus, substituting all the values, we get:
${\text{1}}{\text{.98 cal/gm mol K}} = \dfrac{{8.32{\text{ Joule/gm mol K}}}}{J}$
On simplifying further, we get:
$J = \dfrac{{8.32}}{{1.98}} = 4.20{\text{ J/cal}}$
Thus, the correct option is C.
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 question 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 to get the required answer.
Recently Updated Pages
Difference Between Vapor and Gas: JEE Main 2024

Area of an Octagon Formula - Explanation, and FAQs

Charle's Law Formula - Definition, Derivation and Solved Examples

Central Angle of a Circle Formula - Definition, Theorem and FAQs

Average Force Formula - Magnitude, Solved Examples and FAQs

Boyles Law Formula - Boyles Law Equation | Examples & Definitions

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

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

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

Degree of Dissociation and Its Formula With Solved Example for JEE

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

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

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

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

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

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry
