You supply 100 J of heat to n moles of an ideal gas. In which scenario does the gas temperature rise more?
AConstant volume (isochoric) — all heat goes to raising internal energy
BConstant pressure (isobaric) — the gas can expand and do more work
CBoth scenarios produce the same temperature rise
DConstant pressure — higher Cp means more heat capacity, so more temperature rise per joule
In an isochoric process, W = 0, so all 100 J raises internal energy: ΔT = Q/(nCv). In an isobaric process, some energy is diverted to expansion work (W = nRΔT), leaving less to raise temperature: ΔT = Q/(nCp). Since Cp > Cv, the temperature rise is smaller at constant pressure. Option D is the key misconception: 'more heat capacity' means you need MORE heat per degree, not that you get more temperature rise per joule.
Question 2 Multiple Choice
A sealed rigid container of ideal gas is heated. Which of the following is true?
ANo work is done by the gas on its surroundings
BThe gas does positive work equal to PΔV
CThe heat added exceeds the change in internal energy
DCp is the relevant heat capacity for calculating heat input
Rigid means ΔV = 0, so W = PΔV = 0. By the first law, ΔU = Q, and Q = nCvΔT. Cv (not Cp) applies to constant-volume processes. Option B describes an isobaric process. Option C would require W > 0, since ΔU = Q − W. Option D confuses the two heat capacities: Cp is for constant pressure, Cv for constant volume.
Question 3 True / False
In an isochoric process, the heat added to the gas equals the change in its internal energy.
TTrue
FFalse
Answer: True
W = PΔV = 0 for constant volume. The first law then gives ΔU = Q − W = Q. None of the heat is converted to mechanical work; all of it raises the gas's internal energy (and therefore temperature). This is why Cv is defined via Q = nCvΔT for this process.
Question 4 True / False
Because Cp > Cv, an isobaric process requires less heat input than an isochoric process to achieve the same temperature rise in an ideal gas.
TTrue
FFalse
Answer: False
The opposite is true: Cp > Cv means isobaric heating requires MORE heat to achieve the same ΔT. At constant pressure, some heat is diverted to doing expansion work (W = nRΔT), so extra heat must be supplied. The Mayer relation Cp = Cv + R quantifies this: the extra R per mole per kelvin is exactly the work cost of isobaric expansion.
Question 5 Short Answer
Why does Cp > Cv for an ideal gas, and where does the 'extra' energy go in an isobaric process compared to an isochoric one?
Think about your answer, then reveal below.
Model answer: At constant volume, all heat goes to raising internal energy. At constant pressure, the gas must also expand against external pressure, doing work W = nRΔT. This expansion work leaves the system as mechanical work rather than temperature increase, requiring extra heat input to achieve the same ΔT. Quantitatively, Cp = Cv + R; the R represents the work done per mole per kelvin of temperature rise.
A useful image: heating gas in a sealed rigid cylinder (isochoric) vs. a piston-cylinder that is free to expand (isobaric). The piston must be pushed outward — energy you supplied is doing that pushing rather than warming the gas. So you need more total energy input to reach the same temperature.