A 4.06-L sample of carbon monoxide is collected at 55°C and 0.892 atm. What volume will the gas occupy at 1.05 atm and 20.°C?
A) 1.25 L
B) 3.86 L
C) 3.08 L
D) 4.27 L
E) none of these

Question

Quizplus · Accepted Answer

Use the combined gas law:

(P1V1/T1) = (P2V2/T2)

Convert the temperature to Kelvin:

55°C + 273.15 = 328.15 K

Convert the initial volume to liters:

4.06 L

Plug in the given values:

(0.892 atm)(4.06 L)/(328.15 K) = (1.05 atm)(V2)/(293.15 K)

Solve for V2:

V2 = (0.892 atm)(4.06 L)(293.15 K)/(1.05 atm)(328.15 K)

V2 = 3.077 L

Round to three significant figures:

V2 = 3.08 L

A 406-L Sample of Carbon Monoxide Is Collected at 55°C and and 0.892