On a warm summer day, a large mass of air (atmospheric pressure 1.01×105Pa) is h
ID: 1327396 • Letter: O
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 30.0 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 8.10×104 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.)
T = C
Explanation / Answer
here,
It appears that the air is in thermal equilibrium, and is undergoing an adiabatic expansion.That is, the change in temperature stems solely from its change in volume and pressure.
so adiabatic expansion eqn can be writtern as
(Ti)^(g)*(Pi)^(1-g)= (Tf)^(g)*(Pf)^(1-g)
where
g is gamma which is constant
Ti,Tf are initial and final temp
Pi,Pf are initial and final pressure
(Ti)^(1.4) * (1.05*10^5)^(1-1.4) = (30 + 273.5) ^ (1.4) * (8.10*10^4)^(1-1.4)
0.00980673 Ti^1.4 = 32.4809
Ti^1.4 = 3312.1
Ti = 326.859 K
Ti = 53.709 Degree Celcius
the temperature of the air mass when it has risen to a level at which atmospheric is 53.709 Degree Celcius