
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
152.4k+ 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
JEE Main 2022 (June 29th Shift 2) Maths Question Paper with Answer Key

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 29th Shift 1) Maths Question Paper with Answer Key

JEE Main 2022 (July 26th Shift 2) Chemistry Question Paper with Answer Key

JEE Main 2022 (June 26th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (June 29th Shift 1) Physics Question Paper with Answer Key

Trending doubts
JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Electrical Field of Charged Spherical Shell - JEE

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

Displacement-Time Graph and Velocity-Time Graph for JEE

Collision - Important Concepts and Tips for JEE

Which of the following is the smallest unit of length class 11 physics JEE_Main

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

Charging and Discharging of Capacitor

Brief Information on Alpha, Beta and Gamma Decay - JEE Important Topic

Compressibility Factor Z | Plot of Compressibility Factor Z Vs Pressure for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Laws of Motion Class 11 Notes: CBSE Physics Chapter 4
