At sea level, where the pressure was 104kPa and the temperature 21.1degrees C, a certain mass of air occupied?

Question

2.0m^3. to what volume will the region expand when it has risen to an altitude where the pressure and the temperature are a) 42 kPa, -5.0degrees C b)880Pa, -52.0 degrees C?

Answer 1

start with the perfect gas law:

PV=nRT where P is pressure, V volume, n the number of moles, R a constant, and T is temp

we are comparing two different regions, so we should be prepared to take a ratio...since we are not losing or gaining mass, the value of n, the number of moles of the gas, will be the same...

let's start by solving for V

V=nRT/P

now, V at sea level can be thought of as:

V(sea level)=nRT(sea level)/P(sea level)

V(aloft)=nRT(aloft)/P(aloft)

divide these two equations:

V(aloft)/V(sea level)=

          nRT(aloft)/P(aloft)/[nRT(sea level)/P(sea level)]

after dividing terms we get:

V(aloft)/V(sea level)= T(aloft)/P(aloft)/[T(sea level)/P(sea level)]

or knowing that V(sea level) is 2m^3, we can substitute values and get:

V(aloft)=2m^3[T(aloft)/P(aloft)]/[T(se... level)/P(sea level)]

and now just substitute the appropriate values...keep in mind that since you are comparing temps, you MUST use absolute Kelvin temps...21.1 C = 293.2K

-5C=268.1K

in general, K=C+273.15