Physics and Mathematics
Two Specific Heat of Gas Cp and Cv
1. Concept Overview
In thermodynamics, gases exhibit two distinct specific heats:
- One when pressure is kept constant, called specific heat at constant pressure (Cp).
- Another when volume is kept constant, called specific heat at constant volume (Cv).
This difference arises because when a gas expands at constant pressure, part of the heat supplied does external work against the surrounding pressure, whereas at constant volume, no external work is done.
2. Explanation and Mathematical Derivation
Let:
- [m] = mass of the gas
- [Q] = heat supplied
- [C] = specific heat
- [ΔT] = rise in temperature
Then by definition:
[Q = m C \Delta T]
(a) Specific Heat at Constant Volume (Cv):
When volume is constant,
no work is done ([W = 0]), and the entire heat increases the internal energy ([U]) of the gas.
[Q_v] [= m C_v \Delta T] [= \Delta U]
Therefore,
[C_v] $= \dfrac{1}{m} \left( \dfrac{dU}{dT} \right)_v$
(b) Specific Heat at Constant Pressure (Cp):
When pressure is constant, the gas expands, doing work on the surroundings.
From the first law of thermodynamics:
[Q_p = \Delta U + W]
and since work done by the gas,
[W = p \Delta V]
Then,
[m C_p \Delta T] [= \Delta U + p \Delta V]
Using the ideal gas law [pV = RT] (for 1 mole),
[p \Delta V] [= R \Delta T]
Hence,
[C_p] $= \dfrac{1}{m} \left( \dfrac{dU}{dT} \right)_p + \dfrac{R}{m}$
But from definition of [C_v]:
$\dfrac{1}{m} \left( \dfrac{dU}{dT} \right)_v$ = $C_v
Thus,
[C_p] [= C_v + \dfrac{R}{m}]
For 1 mole of gas:
[C_p – C_v = R]
3. Dimensions and Units
| Quantity | Symbol | S.I. Unit | Dimensions |
|---|---|---|---|
| Specific Heat at Constant Volume | [C_v] | J·kg⁻¹·K⁻¹ or J·mol⁻¹·K⁻¹ | [M⁰L²T⁻²θ⁻¹] |
| Specific Heat at Constant Pressure | [C_p] | J·kg⁻¹·K⁻¹ or J·mol⁻¹·K⁻¹ | [M⁰L²T⁻²θ⁻¹] |
4. Key Features
- [C_p] is always greater than [C_v].
- For an ideal gas, [C_p – C_v = R].
- The ratio of specific heats, denoted by [\gamma = \dfrac{C_p}{C_v}], plays a crucial role in adiabatic processes.
- [C_v] relates to internal energy, while [C_p] includes both internal energy change and work done by expansion.
- These concepts are essential in analyzing heat engines, adiabatic expansion, and isentropic processes.
5. Important Formulas to Remember
| Formula | Description |
|---|---|
| [Q_v = m C_v \Delta T] | Heat supplied at constant volume |
| [Q_p = m C_p \Delta T] | Heat supplied at constant pressure |
| [C_p – C_v = R] | Mayer’s relation |
| [\gamma = \dfrac{C_p}{C_v}] | Ratio of specific heats |
| [C_p = C_v + R] | Relation for 1 mole of an ideal gas |
6. Conceptual Questions with Solutions
1. Why do gases have two specific heats?
Because at constant volume, heat increases only internal energy, while at constant pressure, part of the heat also does external work during expansion.
2. Why is Cp greater than Cv?
At constant pressure, the gas does external work in addition to increasing internal energy. Hence, more heat is required, making [C_p > C_v].
3. What does Cp – Cv = R signify?
It represents the Mayer’s relation, showing that the difference between the two specific heats equals the universal gas constant.
4. What are the units of Cp and Cv in CGS and SI systems?
In SI: J·mol⁻¹·K⁻¹; in CGS: cal·mol⁻¹·°C⁻¹.
5. What does the ratio γ = Cp/Cv indicate?
It determines how a gas behaves in adiabatic processes—higher γ indicates less heat capacity and faster temperature change.
6. Can Cp and Cv be the same?
Only in incompressible substances (like solids and liquids) where no external work is done.
7. For a monatomic ideal gas, what is the value of γ?
[\gamma = \dfrac{5}{3}] or approximately 1.67.
8. For a diatomic ideal gas, what is the value of γ?
[\gamma = \dfrac{7}{5}] or approximately 1.4.
9. What is the physical significance of Cv?
It measures how much heat is required to raise the temperature of a gas without doing external work.
10. Does Cp – Cv = R hold for real gases?
Approximately, yes—when the gas behaves nearly ideally.
7. FAQ / Common Misconceptions
1. Students often think Cp and Cv are constants for all gases.
In reality, Cp and Cv vary slightly with temperature, especially for real gases.
2. Cp and Cv are not always per mole.
They can also be expressed per unit mass or per unit quantity of gas.
3. Cp > Cv does not mean Cp is always large.
Both depend on molecular structure; monatomic gases have smaller values than diatomic ones.
4. The difference Cp – Cv = R applies only to ideal gases.
Correct — real gases show deviations due to intermolecular forces.
5. Cp and Cv cannot be negative.
They are always positive because adding heat increases energy or temperature.
6. Heat supplied at constant volume does not mean pressure is constant.
Pressure changes when volume is fixed and temperature increases.
7. R is not the gas constant for a specific gas.
R is a universal constant equal to [8.314 J·mol⁻¹·K⁻¹].
8. Internal energy depends only on temperature for ideal gases.
Yes, it’s independent of volume or pressure for ideal gases.
9. Cp and Cv are unrelated to degrees of freedom.
Incorrect — they depend directly on molecular degrees of freedom through the equipartition theorem.
10. Specific heats are the same for solids, liquids, and gases.
No — gases have two specific heats because they can expand and do external work.
8. Practice Questions (with Step-by-Step Solutions)
Q1. One mole of an ideal gas has [C_v = 12.5 J·mol⁻¹·K⁻¹]. Calculate [C_p].
Solution:
Using [C_p – C_v = R],
[C_p] [= 12.5 + 8.314 = 20.814 J·mol⁻¹·K⁻¹].
Q2. If [C_p / C_v = 1.4] for air, find [C_v] and [C_p].
Solution:
Let [C_v = x]. Then [C_p = 1.4x].
Using [C_p – C_v = R],
[1.4x – x = 8.314]
[0.4x = 8.314]
[x = 20.785 J·mol⁻¹·K⁻¹].
Hence, [C_p = 29.099 J·mol⁻¹·K⁻¹].
Q3. A gas requires 500 J to raise its temperature by 25°C at constant volume. Find [C_v].
Solution:
[Q = m C_v \Delta T]
[\Rightarrow C_v] [= \dfrac{Q}{m \Delta T}] [= \dfrac{500}{1 \times 25}] [= 20 J·kg⁻¹·K⁻¹].
Q4. For a monatomic ideal gas, show that [\gamma = 5/3].
Solution:
From kinetic theory,
[C_v] [= \dfrac{3}{2}R] [\quad] [C_p] [= \dfrac{5}{2}R].
Hence, [\gamma] [= \dfrac{C_p}{C_v}] [= \dfrac{5/2}{3/2}] [= \dfrac{5}{3}].
Q5. Prove that the heat required at constant pressure is greater than at constant volume for the same temperature rise.
Solution:
Since [C_p – C_v = R > 0],
[C_p > C_v].
Hence, [Q_p] [= m C_p \Delta T] [> Q_v = m C_v \Delta T].
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