# Easy question-but what's wrong?

• bcjochim07
The important phrase is that the air has no time to exchange heat with its surroundings. What does that signify? What relationship between P and T is true in that case?Hmmm... again I forgot that volume changes, and this is an adiabatic process, and volume decreases.

#### bcjochim07

Easy question--but what's wrong?

## Homework Statement

When strong winds rapidly carry air down from mountains to a lower elevation, the air has no time to exchange heat with its surroundings. The air is compressed as the pressure rises, and its temperature can increase dramatically. These warm winds are called Chinook winds in the Rocky Mountains and Santa Ana winds in California. Suppose the air temperature high in the mountains behind Los Angeles is 0 degrees C at an elevation where the air pressure is 60 kPa. What will the air temperature be, in C when the Santa Ana winds have carried this air down to an elevation near sea level where the air pressure is 100 kPa ?

## The Attempt at a Solution

This is a pretty straightforward problem, but for some reason I got the answer wrong.

P1/T1=P2/T2

(60kPa)/(273K) = (100kPa)/(T2) T2=455K = 180 degrees C

The important phrase is that the air has no time to exchange heat with its surroundings. What does that signify? What relationship between P and T is true in that case?

Hmmm... again I forgot that volume changes, and this is an adiabatic process, and volume decreases

For an adiabatic process: P1V1^gamma = P2V2^gamma

Gamma is the specific heat ratio
gamma= Cp/Cv

since air is a mixture of mostly diatomic gases oxygen & nitrogen gamma = 29/20.8 = 1.40

so (60 kPa)(V1)^1.40 = (100kPa)(V2)^1.40
the ratio of V2/V1 = .694

P1V1/T1 = P2V2/T2

(60kPa)(V1)/(273 K) = (100kPa)(V2)/T2
(.220)*T2 = (V2)/(V1) * (100 kPa)
.220T2= 69.4
T2= 314.45 K = 42.45 degrees C

Is this anywhere near how I should approach this problem?

$$P^{\gamma -1} T^{-\gamma} = \mbox{ constant}$$