Question

If the temperature of a 1.5 L reaction vessel containing Argon gas is decreased from 5.60°C to -12.9°C and the pressure remains constant, what is the new volume of gas?

Answer 1

Assuming pressure is constant:

$\frac{V1}{T1}=\frac{V2}{T2}$

T1=5,60 ° С =278.75K

T2=12,9 ° С=260.25K

$\frac{1.5}{278.75}=\frac{V2}{260.25}$

V2=1.4 L