Question

1l of air weighs 1.293g at 0c and 1atm pressure. At what temperature 1l of air at 1atm pressure will weigh 1g?

Answer 1

Answer:

At -61.90 °C 1 liter of air at 1 atm pressure will weigh 1 grams.

Explanation:

Volume of air at 0° C= 1 L

Mass of air =m'= 1.293 g

Molecular mass of air = M

Pressure of the air = 1 atm

Temperature of the air = T = 0° C = 273.15 K

$PV=nRT=\frac{m}{M}RT$

$1 atm\times 1 L=\frac{1.293 g}{M}\times 0.0821 atm L/mol K\times 273.15 K$

M = 17.3438 g/mol

The temperature 1 L of air at 1 atm pressure will weigh 1 gram be T'

m' = 1 gram, T ' = ?, P = 1 atm, V = 1 L

M = 17.3438 g/mol

$PV=\frac{m'}{M}\times RT'$

$T'=\frac{PVM}{R\times m'}=\frac{1atm\times 1 l\times 17.3438 g/mol}{0.0821 atm L/mol K\times 1 g}$

T' = 221.25 K = -61.90 °C

At -61.90 °C 1 liter of air at 1 atm pressure will weigh 1 grams.