Question

why an adiabat is steeper than an isotherm???​

Answer 1

The adiabatic curve is steeper than the isothermal curve, in both the processes of expansion and compression. To reach a same height in a longer distance means lower slope of the line. Hence adiabatic curve is more steeper than isothermal curve.

I assume you refer to curves on a Pressure-Volume graph (PV diagram). There may be a clever non-math way to explain why the dP/dV slope is greater for adiabats than isotherms, and maybe a commenter can do this. Here is my version: the ideal-gas law, for an isothermal process, yields PV = constant. Thus the pressure is inversely proportional to the volume. For an ADIABATIC process, PV^(gamma) = constant, where gamma (the “adiabatic constant”) is the ratio of two specific heats and is ALWAYS GREATER THAN ONE. (As to why this is so, see any intro calculus-based physics textbook; it’s about a one-page derivation.) The result is that, for the adiabatic process, the pressure is inversely proportional to the volume raised to a power greater than one - a stronger inverse dependence than for the isothermal case, which implies that pressure falls more rapidly as volume increases.

Answer 2

Answer:

The adiabatic curve is steeper than the isothermal curve, in both the processes of expansion and compression. To reach a same height in a longer distance means lower slope of the line. Hence adiabatic curve is more steeper than isothermal curve.

Explanation:

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