The mass of 1 m" of dry air at standard temperature and pressure is 1.29 kg;
therefore the mass of dry air in a room 8 x 6 x 4 m, at an atmospheric pressure of
96 kPa and a temperature of 20°C, is:
400410
A 195 kg
B 201 kg
C 206 kg
D 219 kg

if anyone can answer this rubic cube of a question, please include the calculations
cheers

The mass of 1 cubic meter (?) of air is 129 kg. The mass of air is proportional to its volume.

The "combined gas law" (combining Boyle's, Charles, and Gay-Lussac's laws) is:

\Large \frac {P_1V_1}{T_1}\,=\,\frac {P_2V_2}{T_2}

at STP,
P1 = 1 atm = 101.32500 KPa
P2 = 96 KPa
V1 is unknown for the volume specified.
V2 is given, 8 x 6 x 4 m = 192 m^3
T1 at STP is 273.15 Kelvins (0 °C).
T2 is 293.15 Kelvins (20 °C)

Solve the equation for V1 (the volume at STP). Set up a ratio equation wherein 1 cubic meter is to 129 Kg as V2 cubic meters is to the unknown weight. Solve for the unknown weight.

I've set it up for you. You can do the math.

This should have been put under Chemistry or possibly Physics.

The mass of 1 cubic meter (?) of air is 129 kg. The mass of air is proportional to its volume.

The "combined gas law" (combining Boyle's, Charles, and Gay-Lussac's laws) is:

\Large \frac {P_1V_1}{T_1}\,=\,\frac {P_2V_2}{T_2}

at STP,
P1 = 1 atm = 101.32500 KPa
P2 = 96 KPa
V1 is unknown for the volume specified.
V2 is given, 8 x 6 x 4 m = 192 m^3
T1 at STP is 273.15 Kelvins (0 °C).
T2 is 293.15 Kelvins (20 °C)

Solve the equation for V1 (the volume at STP). Set up a ratio equation wherein 1 cubic meter is to 129 Kg as V2 cubic meters is to the unknown weight. Solve for the unknown weight.

I've set it up for you. You can do the math.

This should have been put under Chemistry or possibly Physics.

Perito
thanks for your help, I've still not figured out the answer yet, but I'm sure I will.
thanks again
peter

D.) 219 = 1.29 x 273/293 x 96/101 x 192