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

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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?

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  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

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