A sample of argon has a volume of 1.2 L at STP. If the temperature is

increased to 22°C and the pressure is lowered to 0.80 atm, what will

the new volume be, in L?

the new volume would be 1.62 L

Explanation:

stp is 0 degrees C or 273.15 K and 1atm for the pressure

22 degrees C is 295.15 K

use the combined gas law formula and solve for v2

v2=p1 (v1) (t2) / p2 (t1)

v2=1atm (1.2L) (295.15) / 0.8atm (273.15K)

v2=1.62