Physics 2nd Paper · Classes 11-12 · Chapter 1 · 1.4
Molar Specific Heat.
Take one gas and warm it by the same 1 K, and it can still ask for two different amounts of heat: less in a sealed box, more under a free piston. Where does the extra go? Into work, whenever the gas is allowed to expand.
1. Molar specific heat
The heat needed to raise the temperature of 1 mole of a gas by 1 degree is its molar heat capacity, or molar specific heat. The symbol is C.
If ΔQ of heat raises the temperature of n moles of gas by ΔT, then C = ΔQ/(nΔT) (1.11).
Measure ΔQ in joules, n in moles and ΔT in kelvin, and the unit of C comes out as J mol⁻¹ K⁻¹.
A gas has two specific heats, so it has two molar specific heats as well: CP at constant pressure and CV at constant volume.
2. Why CP is larger than CV
CP is the heat needed to raise 1 mole of gas by 1 K at constant pressure. By definition CP = ΔQ/(nΔT) (1.12).
CV is the heat needed to raise 1 mole of gas by 1 K at constant volume. By definition CV = ΔQ/(nΔT) (1.13).
Experiments show that CP is greater than CV. Why? At constant volume the gas does no external work, so all the heat goes into raising pressure and temperature. At constant pressure the heat raises the temperature and also pays for the gas pushing outwards against the external pressure as it expands.
So CP = CV + (the heat equivalent of the work against the external pressure), hence CP > CV. Warm the gas by 1 K in both set-ups and compare how the heat splits.
Why the extra heat
The drawing is by hand and the split comes from the formula, not from a measurement of yours.
3. CP − CV = R
The proof is short. Heat 1 mole of gas at constant P so that its temperature rises by dT (take dT = 1 K). The CV dT part of the heat raises the temperature. As the piston moves by x the work is P × A × x = P dV. Hence CP = CV + P dV (1.14).
Differentiate PV = RT (1.15) at constant pressure and you get P dV = R dT = R, because dT = 1 K. Put that into (1.14) and CP = CV + R.
So for an ideal gas the difference of the two molar specific heats equals the gas constant: CP − CV = R (1.16).
R is positive, so CP > CV. With R = 8.314 J K⁻¹ mol⁻¹, CP − CV = 8.314 J K⁻¹ mol⁻¹.
Now γ: the ratio of the specific heat at constant pressure to that at constant volume is γ = CP/CV. Divide (1.16) by CV and CP/CV − 1 = R/CV, that is γ − 1 = R/CV. So CV = R/(γ − 1) [1.16(a)].
4. Gas type and γ
A monatomic gas has CV = (3/2)R, CP = (5/2)R and γ = 5/3 = 1.67.
A diatomic gas has CV = (5/2)R, CP = (7/2)R and γ = 7/5 = 1.40.
A polyatomic gas has CV = 3R, CP = 4R and γ = 4/3 = 1.33. In every row CP − CV = R holds, check it yourself.
γ by gas type
Which gas goes in which group? Monatomic: He, Ne, Ar (γ = 1.67). Diatomic: H₂, O₂, N₂, Cl₂ (γ = 1.40). Triatomic or polyatomic: CO₂, C₂H₆, NH₃ (γ = 1.33). The book heads the last group polyatomic but lists it as triatomic; it is one and the same group. Drop the gases and the γ values into their boxes.
Sort the gases
γ is fixed for gases of the same molecular structure and differs for gases of different structure.
Why does γ matter? (a) Knowing γ tells how many atoms each molecule of the gas has. (b) The speed of sound in a gas depends on γ. (c) γ is needed to study the adiabatic process of a gas.
5. Working from γ
Example 1.5: for a polyatomic gas with γ = 1.33 and R = 8.31, CV = R/(γ − 1) = 8.31/0.33 = 25.18 J mol⁻¹ K⁻¹ and CP = CV + R = 25.18 + 8.31 = 33.49 J mol⁻¹ K⁻¹. Put your own γ into the calculator below too.
Molar heats from γ
R is taken as 8.31; the numbers are for practice, not a measurement of yours.
It works backwards too. If γ = 5/3 then CV = R/(γ − 1) = R/(2/3) = (3/2)R and CP = (5/2)R, the values of a monatomic gas.
So γ tells you the kind of gas: 1.67 means monatomic, 1.41 or 1.40 means diatomic, 1.33 means triatomic or polyatomic.
The adiabatic slope is −γP/V, so the larger γ is, the steeper the curve. A monatomic gas (helium) has the largest γ (1.67), so its adiabatic curve is the steepest.
Heat sums come from the same relation. Heating n moles by ΔT needs Q = n CV ΔT at constant volume and Q = n CP ΔT at constant pressure. For example 2 mol of a diatomic gas with ΔT = 10 K needs Q = 2 × (5/2)(8.31) × 10 ≈ 416 J at constant volume.
How much heat to warm it
R is taken as 8.31; the numbers are for practice, not a measurement of yours.
6. Check yourself
Three questions in the exam's style.
Question 1
The ratio of the two molar specific heats of a gas is a constant quantity. This constant is expressed by the symbol:
Put the relations to work in the second.
Question 2
For a gas whose adiabatic process gives PV^1.33 = constant:
i. C_P = 1.33 C_V
ii. the isothermal curve is 1.33 times as steep as the adiabatic curve
iii. the gas is made of triatomic or polyatomic molecules
In the last, match γ to the kind of gas.
Question 3
What is the value of γ for CO₂ gas?
The whole note on eight cards.
Eight cards
Source: Physics 2nd Paper, Classes 11-12 (NCTB), Chapter 1 “Thermodynamics”, section 1.4.