In: Chemistry
A sample of oxygen gas occupies a volume of 9.30 L at 52.0°C and 0.790 atm. If it is desired to increase the volume of the gas sample to 11.0 L, while decreasing its pressure to 0.569 atm, the temperature of the gas sample at the new volume and pressure must be °C.
Initial Final
V1 = 9.30L V2 = 11L
P1 = 0.790atm P2 = 0.569atm
T1 = 52+273 = 325K T2 =
P1V1/T1 = P2V2/T2
T2 = P2V2T1/P1V1
= 0.569*11*325/(0.790*9.30)
= 276.87K
= 276.87-273 = 3.870C >>>>answer