19-7 The Molar Specific Heats of an Ideal Gas#
Prompts
For a monatomic ideal gas, write \(E_{\text{int}}\) in terms of \(n\) and \(T\). Why does \(E_{\text{int}}\) depend only on \(T\)?
Define molar specific heat at constant volume \(C_V\). Why is \(C_V = \frac{3}{2}R\) for a monatomic gas?
Why is \(C_P > C_V\)? Derive or state \(C_P = C_V + R\).
For a given \(\Delta T\), is \(\Delta E_{\text{int}}\) the same for a constant-volume process and a constant-pressure process? Why?
Lecture Notes#
Overview#
Molar specific heat is heat per mole per degree: \(Q = nC\,\Delta T\).
\(C_V\) (constant volume): heat goes entirely into internal energy; \(W = 0\).
\(C_P\) (constant pressure): heat goes into internal energy and work; \(C_P = C_V + R\).
\(\Delta E_{\text{int}} = n C_V \Delta T\) for any process—path independent.
Internal energy of a monatomic ideal gas#
For a monatomic gas (He, Ne, Ar), internal energy is the sum of translational kinetic energies. With \(K_{\text{avg}} = \frac{3}{2}kT\) per molecule:
\(E_{\text{int}}\) depends only on \(T\), not on \(p\) or \(V\).
For any ideal gas: \(E_{\text{int}} = n C_V T\); \(\Delta E_{\text{int}} = n C_V \Delta T\) (any process).
Molar specific heat at constant volume#
At constant volume, \(W = 0\) → \(\Delta E_{\text{int}} = Q\) (first law).
So \(C_V = \Delta E_{\text{int}}/(n\,\Delta T)\).
Monatomic gas: \(\Delta E_{\text{int}} = \frac{3}{2}nR\,\Delta T\) → \(C_V = \frac{3}{2}R \approx 12.5\) J/(mol·K).
Molar specific heat at constant pressure#
At constant pressure, the gas expands: \(W = p\,\Delta V = nR\,\Delta T\). From the first law \(\Delta E_{\text{int}} = Q - W\):
\(C_P > C_V\) because heat must supply both the increase in \(E_{\text{int}}\) and the work done by the gas.
Gas type |
\(C_V\) |
\(C_P\) |
|---|---|---|
Monatomic |
\(\frac{3}{2}R\) |
\(\frac{5}{2}R\) |
Diatomic |
\(\frac{5}{2}R\) |
\(\frac{7}{2}R\) |
Polyatomic |
\(3R\) |
\(4R\) |
Path independence of \(\Delta E_{\text{int}}\)
For an ideal gas, \(\Delta E_{\text{int}} = n C_V \Delta T\) for any process between two temperatures. It does not depend on whether the path is constant volume, constant pressure, or something else.
Poll: Which path delivers more heat?
Two copies of the same amount of the same ideal gas undergo processes A (isochoric) and B (isobaric) from the same initial to the same final state on a \(p\)–\(V\) diagram. Which gas had more heat delivered to it?
[FIGURE: pV diagram with two paths from same \((p_i,V_i)\) to same \((p_f,V_f)\)—vertical isochoric path A, horizontal isobaric path B]
(A) Gas A
(B) Gas B
(C) Same
(D) Cannot tell
Poll: Specific heat
To raise the temperature of 1 mol of ideal gas by 10 K at constant pressure, you need ___ heat than at constant volume.
(A) More
(B) Less
(C) The same
Example: Heat at constant pressure—find \(C_P\), DoF, and work#
Example: Heat at constant pressure—find \(C_P\), DoF, and work
You add 75 J of heat at constant pressure to 1 mol of gas, and the temperature increases by exactly 2.00 K. Find (a) \(C_P\) and the number of degrees of freedom, (b) the work done by the gas.
Solution: (a) \(C_P = Q/(n\,\Delta T) = 37.5\) J/(mol·K) \(\approx \frac{5}{2}R\) → diatomic, \(f = 5\). (b) \(W = p\,\Delta V = nR\,\Delta T = 16.6\) J; or \(W = Q - \Delta E_{\text{int}} = Q - n C_V \Delta T\).
Summary#
\(E_{\text{int}} = \frac{3}{2}nRT\) (monatomic); \(\Delta E_{\text{int}} = n C_V \Delta T\) (any ideal gas, any process).
\(Q = n C_V \Delta T\) (constant \(V\)); \(Q = n C_P \Delta T\) (constant \(p\)).
\(C_P = C_V + R\)—constant pressure requires extra heat for expansion work.