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November 23, 2014

November 23, 2014

Posted by **Kelly** on Wednesday, June 23, 2010 at 9:02pm.

- Thermal physics -
**MathMate**, Thursday, June 24, 2010 at 8:46amUse PV=nRT

where

P=pressure

V=volume

T=temperature in Kelvin (Celsius+273.15°)

n=number of moles (quantity of substance)

R=gas constant

In the given case, the quantity of air remain the same, so n is a constant. The pressure is assumed constant, so the equation is reduced to

V1/T1=V2/T2

where 1 and 2 represent initial and final states.

V1=0.6L

T1=273.15+24°

T2=273.15+37°

V2=(V1/T1)*T2

=0.6L*(310.15/297.15)

At 0°C, air weighs 1.29 kg/m².

So the volume of 0.6L at 24° has to be transformed to the volume at 0° for the calculation of mass.

Also,

1.29 kg-m^{-3}

=1.29 g-l^{-1}

Volume at 0°

=(V1/T1)*T2

=0.6L*(273.15/297.15)

=0.552L

Mass of air = 0.552*1.29 g

=0.712 g

Note: the given answer (7.74*10^-4 kg) for the mass assumes that air weighs 1.29 kg/m^3 at 24°C.

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