What pressure, in atm, is required to reduce 60.0 mL of a gas at standard conditions to
10.0 mL at a temperature of 25.0 °C?

Respuesta :

P1=1atm
V1=60mL
V2=10mL
P2=?
P2=P1*V1/V2
=1*60/10
=6 atm#