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Heat Capacity - Relationship between Cp and Cv MCQ - Practice Questions with Answers

Edited By admin | Updated on Sep 25, 2023 25:23 PM | #NEET

Syllabus

Quick Facts

Heat Capacity is considered one the most difficult concept.
13 Questions around this concept.

Solve by difficulty

MEDIUM

Helium gas goes through a cycle ABCDA ( consisting of two isochoric and two isobaric lines ) as shown in the figure. The efficiency of this cycle is nearly: ( Assume the gas to be close to ideal gas )

15.4

9.1

10.5

12.5

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Concepts Covered - 2

Heat Capacity

Heat Capacity

The heat capacity of a system is defined as "The quantity of heat required for increasing the temperature of one mole of a system through 1⁰C". It is given as follows:

$\mathrm{C}=\frac{\mathrm{dq}}{\mathrm{dT}} \quad \ldots .(1)$

(i) Heat capacity at constant volume

According to first law of thermodynamics,

$\begin{array}{l} { \mathrm{dq}=\mathrm{dE}+\mathrm{PdV}} \\\\ {\text { On substituting the value of } \mathrm{dq} \text { in equation }(2)} \\\\ {\mathrm{C}=\frac{\mathrm{dE}+\mathrm{PdV}}{\mathrm{dT}}.......(3)} \\\\ \textrm{If volume is constant then }\\\\ {\mathrm{C}_{\mathrm{v}}=(\mathrm{d} \mathrm{E} / \mathrm{d} \mathrm{T})_{\mathrm{v}} \quad \ldots . .(4)}\end{array}$

Hence the heat capacity at constant volume of a given system may be defined as the rate of change of internal energy with temperature.

(ii) Heat capacity at constant pressure

If pressure is constant, equation (3) becomes as follows:

$\mathrm{ C_{P} =\frac{d E+P d V}{d T} } \\ \\ \text { or } \quad \mathrm{C_{P} =(dq / d T)_{P}= \frac{dH}{dT}\ .\ .\ .\ .\ .\ . (5)}$

Hence the heat capacity at constant pressure of a system may be defined as the rate of change of enthalpy with temperature.

For the proof of equation (5), we have to learn the relation between Enthalpy (H) and Internal energy (E)

Relation between Enthalpy and Internal Energy

Enthalpy (H) and Internal energy (E) are related as

$\mathrm{H = E + PV}$

$\therefore \mathrm{dH = dE +d(PV)}$

$\therefore \mathrm{dH = dE +PdV + VdP}$

At constant pressure, dP =0

$\mathrm{dH = dE +PdV=(dq)_p}$

Hence, the heat supplied at constant pressure is equal to the Enthalpy

Relation Between Cp And Cv

For one mole of a gas C_p and C_v are known as molar heat capacities and the difference between them is equal to the work done by one mole of gas in expansion on heating it through 1^oC.

We know that

$\mathrm{H = U + PV}$

$\Rightarrow \mathrm{dH = dU +d(PV)}$

$\Rightarrow \mathrm{n C_p dT = n C_vdT +nR\ dT\ [\because PV= nRT]}$

$\Rightarrow \mathrm{C_p -C_v= R}$

Other Relation between C_p and C_v

$\gamma = \frac{C_P}{C_V}$

Now, C_V and C_P can be represented as

$\mathrm{C_V= \frac{f\ R}{2} and\ C_P= (\frac{f}{2}+1) R}$

where f is the degree of freedom

(1) For Monoatomic Gas:

$f= 3, C_V = \frac{3R}{2}, C_P=\frac{5R}{2}, \gamma=\frac{5}{3}$

(2) For Diatomic Gas:

$f= 5, C_V = \frac{5R}{2}, C_P=\frac{7R}{2}, \gamma=\frac{7}{5}$

(3) For Polyatomic Gas:

$f= 6, C_V = 3R, C_P=4R, \gamma=\frac{4}{3}$

Note: These degree of freedom values do not include the Vibrational degree of freedom. We have to neglect the vibrational degree of freedom, unless mentioned otherwise.

Remember these relation between C_p and C_v for calculation in questions