Question

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 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.)

Answer 1

`T_(2)=278.80 K`

Step-by-step explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

`((V_(1))/(V_(2)))^(\gamma -1) = (T_(2))/(T_(1))`.

Now, let's use the ideal gas equation to the initial and the final state:

`(p_(1) V_(1))/(T_(1)) = (p_(2) V_(2))/(T_(2))`

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

`(V_(1))/(V_(2))=(T_(1)p_(2))/(T_(2)p_(1))`

Combining this equation with the first equation we have:

`((T_(1)p_(2))/(T_(2)p_(1)))^(\gamma -1) = (T_(2))/(T_(1))`

`((p_(2))/(p_(1)))^(\gamma -1) = (T_(2)^(\gamma))/(T_(1)^(\gamma))`

Now, we just need to solve this equation for T₂.

`T_(1)\cdot ((p_(2))/(p_(1)))^{(\gamma - 1)/(\gamma)} = T_(2)`

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

`p_(2)=8.00\cdot 10^(4) Pa \\p_(1)=1.01\cdot 10^(5) Pa\\ T_(1)=298 K\\ \gamma=1.40`

Finally, T2 will be:

`T_(2)=278.80 K`