On a warm summer day, a large mass of air (atmospheric pressure 1.01×105Pa) is heated by the ground to a temperature of 26.0 ∘C and then begins to rise through the cooler surrounding air.
1.Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.60×104 Pa . Assume that air is an ideal gas, with γ=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.)
Adiabatic Process is the one in which no heat is gained or lost by the system. Thus given system of air can be considered as undergoing the adiabatic process.
then we'll have
Now Using Ideal Gas Equation: PV= nRT
where n= no. of moles of gas,
R= Universal Gas Constant
T= Temperature.
Then we'll have
(
because n and R are the constants for same
system).....................(1)
Now Initial Temperatue,
Initial Pressure,
Final Pressure,
Let Final Temperature be
Given
Using above in equation 1, we'll get
(ANS)
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