Question

A gas is 90.82L at 29.8 atm and 99.9°C. What volume would the gas be at standard temperature and pressure.

Answer 1

1981.38L

Step-by-step explanation:

P=29.8atm;

T=273+99.9=372.9⁰K

V=90.82L

P₀=1atm

T₀=273⁰K

V₀-?

PV/T=P₀V₀/T₀

V₀=PVT₀/P₀T=

`= (29.8 * 90.82 * 273)/(1 * 372.9) = 1981.38l`