On a warm summer day, a large mass of air (atmospheric pressure 1.01×105Pa) is h

Question

On a warm summer day, a large mass of air (atmospheric pressure 1.01×105Pa) is heated by the ground to a temperature of 28.0 C and then begins to rise through the cooler surrounding air.

Part A

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.50×104 Pa . Assume that air is an ideal gas, with =1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C per 100 m of altitude, is called the dry adiabatic lapse rate.)

On a warm summer day, a large mass of air (atmospheric pressure 1.01×105Pa) is heated by the ground to a temperature of 28.0 C and then begins to rise through the cooler surrounding air.

Part A

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.50×104 Pa . Assume that air is an ideal gas, with =1.40. (This rate of cooling for dry, rising air, corresponding to roughly 1 C per 100 m of altitude, is called the dry adiabatic lapse rate.)

Explanation / Answer

Since this is a adiabatic process

P1^(1-gamma)*T1^gamma = P2^(1-gamma)*T2^gamma

(1.01*10^5)^(1-1.4)*(273+28)^1.4 = (8.5*10^4)^(1-1.4)*T2^1.4

T2 = 286.52 K

T2 = 286.52 - 273 = 13.52 deg C

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