For an ideal gas, the relation between molar heat capacities is $C_p - C_v = R$. Why is $C_p$ always greater than $C_v$?
AAt constant pressure, extra heat goes into doing work as the gas expands, in addition to raising temperature
BMolecules collide more often at constant pressure
CSpecific heats depend only on the gas type, not on the process
DThe gas absorbs less heat at constant pressure
Answer & Solution
Correct answer: A. At constant pressure, extra heat goes into doing work as the gas expands, in addition to raising temperature
At constant volume, all the heat $Q = nC_v \Delta T$ raises internal energy. At constant pressure the gas expands, so part of the heat $Q = nC_p \Delta T$ goes into work done on the surroundings ($PdV$), and only the rest raises the temperature. To produce the same $\Delta T$ at constant pressure, the system has to be given more heat. Hence $C_p > C_v$.
Using the first law for an ideal gas: $nC_p \Delta T = nC_v \Delta T + nR \Delta T$, which rearranges directly to $C_p - C_v = R$ (Mayer's relation).
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