Physics · Thermodynamics · NEET
Ordinary (or specific) heat capacity s is the heat needed to raise 1 kilogram of a substance by 1 kelvin, with unit J per kg per K. Molar specific heat C is the heat needed to raise 1 mole of the substance by 1 kelvin, with unit J per mol per K. For gases we almost always use the molar version because a mole is a fixed number of molecules, so C tells us about each molecule's behaviour, not about how many grams we happen to have. NCERT writes molar heat as Q = mu C delta T, where mu is the number of moles.
When you heat a gas you must state whether the volume is held fixed or the pressure is held fixed. At constant volume the gas does no work, so all heat goes into internal energy: Cv = (delta U / delta T). At constant pressure the gas expands and does work P delta V, so it needs extra heat: Cp = Cv + R. A solid barely expands, so its two values are almost equal and we usually quote just one. This is why gases always come with the pair Cv and Cp.
Degrees of freedom f is the number of independent ways a molecule can store energy: 3 translations for a point atom, plus 2 rotations for a linear (diatomic) molecule, plus vibration at high temperature. The law of equipartition gives each degree (1/2)R of molar energy, so internal energy of one mole is U = (f/2)RT. Since Cv = dU/dT, we get Cv = (f/2)R. So counting f directly gives Cv, then Cp = Cv + R and gamma = 1 + 2/f follow.
Monatomic gas (f = 3, like He, Ar): Cv = (3/2)R, Cp = (5/2)R, gamma = 5/3 = 1.67. Diatomic gas (f = 5, like O2, N2, H2 at room temperature): Cv = (5/2)R, Cp = (7/2)R, gamma = 7/5 = 1.4. Triatomic or polyatomic (non-linear, f = 6): Cv = 3R, Cp = 4R, gamma = 4/3 = 1.33. Also remember Cp is always greater than Cv, and Cp - Cv = R for every ideal gas.
For an isobaric (constant pressure) process, work by the gas is W = P delta V = mu R delta T, and heat absorbed is Q = mu Cp delta T. Dividing, W/Q = R/Cp. For a monatomic gas Cp = (5/2)R, so W/Q = R divided by (5/2)R = 2/5. This exact result was asked in NEET 2018, which is why the degrees-of-freedom values feed directly into thermodynamics numericals.
The volume (V) of a monatomic gas varies with its temperature (T) as a straight line through the origin (an isobaric process from A to B). The ratio of the work done by the gas to the heat absorbed by it, when it goes from A to B, is
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Cv = (f/2)R and Cp = (f/2 + 1)R, where f is the number of degrees of freedom and R = 8.31 J per mol per K. The ratio gamma = Cp/Cv = 1 + 2/f.
At constant volume no work is done, so heat only raises internal energy. At constant pressure the gas also expands and does work P delta V, needing extra heat equal to R per mole. Hence Cp = Cv + R, so Cp is larger by exactly R.
A diatomic gas at room temperature has f = 5, so Cv = (5/2)R, Cp = (7/2)R, and gamma = 7/5 = 1.4.
It is exact only for an ideal gas. For real gases it is a close approximation at ordinary pressures and temperatures. For NEET, treat gases as ideal unless told otherwise, so Cp - Cv = R applies.
No. Molar specific heat is defined per mole, so it depends on the nature of the gas (its degrees of freedom) and the process, but not on how many moles you have.
