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Heat Capacity - Relationship between Cp and Cv - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

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  • Heat Capacity is considered one the most difficult concept.

  • 25 Questions around this concept.

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QuestionEASY

A certain reaction has a heat capacity \mathrm{\left(C_p\right)} of  \mathrm{50 \mathrm{~J} / \mathrm{K}}. If \mathrm{500 \mathrm{~J}} heat is added to the system, calculate the resulting temperature change \mathrm{(\Delta T)} of the reaction.

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Consider a system that consists of 200 g of water initially at 25°C. The system absorbs 1500 J of heat from the surroundings. Calculate the final temperature of the water assuming no phase change occurs.

Given:
Specific heat capacity of water (C or c) =4.18J/g\degree C

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Heat Transfer and Temperature Change

A piece of copper weighing 500 \mathrm{~g} is heated until its temperature increases from \mathrm{20^{\circ} \mathrm{C} \: to \: 80^{\circ} \mathrm{C}}. Calculate the amount of heat absorbed by the copper. Given: Specific heat of copper \mathrm{c=0.387 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}.}
 

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Consider a 3-mole sample of an ideal monatomic gas initially at a temperature of 200 K and a volume of 10 liters. The gas undergoes an isochoric process, during which it absorbs 1500 J of heat. If the final temperature of the gas is 400 K, calculate the heat capacity at constant volume \mathrm{(C_v)} for this gas.

Given: Initial temperature \mathrm{(T_1)} = 200 K
Initial volume \mathrm{(V_1)} = 10 liters
Heat absorbed (Q) = 1500 J
Final temperature \mathrm{(V_2)} = 400 K
The gas is ideal and monatomic.

View Solution

A 3-mole sample of an ideal monatomic gas is initially at a temperature of 200 K and a volume of 10 liters. The gas undergoes an isochoric process, during which it absorbs 1500 J of heat. If the final temperature of the gas is 400 K, calculate the heat capacity at constant volume \mathrm{(C_v)} for this gas.
 

Given: Initial temperature \mathrm{(T_1)} = 200 K
Initial volume \mathrm{(V_1)} = 10 liters
Heat absorbed(Q) = 1500 J
Final temperature \mathrm{(T_2)} = 400 K
The gas is ideal and monatomic.

View Solution

A sample of an ideal diatomic gas is initially at a temperature of 500 K and a volume of 4 liters. The gas undergoes an isochoric process, during which it absorbs 1800 J of heat. If the final temperature of the gas is 700 K, calculate the heat capacity at constant volume \mathrm{(C_{v})} for this gas.

Given: Initial temperature \mathrm{(T_{1})} = 500 K
Initial volume \mathrm{(V_{1})} = 4 liters
Heat absorbed (Q) = 1800 J
Final temperature \mathrm{(T_{2})} = 700 K
The gas is ideal and diatomic.

View Solution

Calculate the heat required to raise the temperature of 100 g of water from \mathrm{25^{\circ} \mathrm{C}} to its boiling point at
\mathrm{100^{\circ} \mathrm{C}}. The specific heat capacity of water is \mathrm{4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)}

View Solution
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The heat of fusion for a certain substance is \mathrm{120 \mathrm{~J} / \mathrm{g} \text {. }} How much heat is required to completely melt 50 grams
of this substance from solid to liquid at its melting point?

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A chemical reaction takes place in a bomb calorimeter. The calorimeter is surrounded by a water bath, and the temperature of the water bath increases from \mathrm{20^{\circ} \mathrm{C} \: to \: 22^{\circ} \mathrm{C}} upon the reaction. The heat capacity of the water bath is \mathrm{1500 \mathrm{~J} /{ }^{\circ} \mathrm{C}}. Calculate the heat released \mathrm{(\Delta H)} by the reaction.
 

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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 Cp and Cv 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 1oC.

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 Cp and Cv 

\gamma = \frac{C_P}{C_V}

Now, CV and CP 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 Cp and Cv for calculation in questions

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Heat Capacity
Relation Between Cp And Cv

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Heat Capacity - Relationship between Cp and Cv Current Topic

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Introduction To Heat, Internal Energy And Work
Isothermal Expansion of an Ideal Gas
Graphical Comparison of Thermodynamic Processes
Heat Capacity - Relationship between Cp and Cv
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Reference Books

Heat Capacity

Chemistry Part I Textbook for Class XI

Page No. : 168

Line : 40

Relation Between Cp And Cv

Chemistry Part I Textbook for Class XI

Page No. : 169

Line : 10

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