In: Chemistry
A sample of an ideal gas has a volume of 3.65 L at 13.00 °C and 1.80 atm. What is the volume of the gas at 24.00 °C and 0.990 atm?
Initial conditions of ags:
Pressure P1 = 1.80 atm
Volume V1 = 3.65 L
Temperature T1 = 13 oC = 13 + 273 K = 286 K
Final conditions of gas:
Pressure P2 = 0.990 mmHg
Volume V2 = ? L
Temperature T2 = 24 oC = 24 + 273 K = 297 K
We know that P1V1/T1 = P2V2/T2
V2 = (P1V1/T1) (T2/P2)
= ( 1.80 atm x 3.65 L/ 286 K ) ( 297 K/ 0.990 atm)
= 6.89 L
V2 = 6.89 L
Therefore, volume of the gas at 24.00 °C and 0.990 atm = 6.89 L