Suppose you give an ideal gas some amount of heat, Q, in one of two ways: one is to keep the gas at constant pressure, so you allow it to expand (and therefore do some work) in the process. The other is to keep the gas at constant volume, so no work is done on or by it. In which case will the ratio Q/ΔT be greater, and why?

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Chemistry

13 October, 21:43

Suppose you give an ideal gas some amount of heat, Q, in one of two ways: one is to keep the gas at constant pressure, so you allow it to expand (and therefore do some work) in the process. The other is to keep the gas at constant volume, so no work is done on or by it. In which case will the ratio Q/ΔT be greater, and why?

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Karla Gaines · Answer 1

Q/ΔT at constant pressure process will be greater than at constant volume process.

Explanation:

In constant pressure process, the heat given to the ideal gas does both that performs some work (expansion) and increases its internal energy also.

Q = ΔU+W

Q = heat supplied

ΔU = change in internal energy

W = work done.

Whereas in constant volume process, the heat given to the ideal gas only increases the internal energy of the gas and does no work.

therefore, ΔT for constant volume process will be higher than that of in constant pressure process.

⇒Q/ΔT at constant pressure process will be greater than at constant volume process.