Chemistry · Thermodynamics · NEET
At constant volume (Cv), all the heat you add goes only to raising the temperature (increasing internal energy U). At constant pressure (Cp), the gas expands, so part of the heat also does work pushing the surroundings back. So you must supply extra heat to get the same temperature rise. That extra amount is exactly R per mole. This is why Cp is always bigger than Cv.
Start from H = U + pV. For one mole of an ideal gas, pV = RT. So H = U + RT. When temperature changes by ΔT, ΔH = ΔU + R ΔT. Now use qV = Cv ΔT = ΔU and qp = Cp ΔT = ΔH. Putting these in: Cp ΔT = Cv ΔT + R ΔT. Cancel ΔT and you get Cp - Cv = R. The R is simply the gas constant that appears because pV = RT.
The clean form Cp - Cv = R is for ONE mole (molar heat capacities). For n moles, the total heat capacities give Cp - Cv = nR, because pV = nRT. In NEET, if it says 'molar' or 'one mole', use R. If it gives n moles of total heat capacity, use nR. Always check the wording.
No. This exact result is only for an IDEAL gas, because we used pV = RT. For real gases the difference is not exactly R. For solids and liquids, volume barely changes on heating, so Cp and Cv are almost equal (their difference is very small). NEET questions on this relation always assume an ideal gas.
qV is the heat at constant volume; here no work is done, so all heat becomes ΔU. So qV = Cv ΔT = ΔU. qp is the heat at constant pressure; here the gas expands and the heat becomes ΔH. So qp = Cp ΔT = ΔH. The whole derivation just plugs these two facts into ΔH = ΔU + R ΔT.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Mayer's relation says that for one mole of an ideal gas, the molar heat capacity at constant pressure minus the molar heat capacity at constant volume equals the gas constant: Cp - Cv = R.
R is the universal gas constant, equal to 8.314 J K⁻¹ mol⁻¹ (or about 2 cal K⁻¹ mol⁻¹). So for an ideal gas, Cp is about 8.314 J K⁻¹ mol⁻¹ larger than Cv.
We take one mole so that pV = RT (instead of pV = nRT). This makes Δ(pV) = R ΔT, which directly gives the clean result Cp - Cv = R. For n moles the same steps give Cp - Cv = nR.
Yes. The ratio γ = Cp/Cv is used in adiabatic processes. Since Cp = Cv + R, once you know Cv (for example (3/2)R for a monatomic gas), you can find Cp and then γ. This links the relation to adiabatic NEET problems.
Yes. It is from Class 11 Chemistry Thermodynamics (and also appears in Physics kinetic theory). NEET can ask the derivation, the meaning of R here, or use it inside numerical problems on heat capacity and adiabatic processes.
