A rubber balloon was filled with helium at 25˚C and placed in a beaker of
liquid nitrogen at -196˚C. The volume of the cold helium was 15 cm3 Assuming ideal gas behavior and isobaric conditions, what was the volume
of the helium at 25˚C?
Isobaric means constant pressure.
V1/T1) = (V2/T2)
Remember T must be in kelvi.
To solve this problem, we can use the ideal gas law, which states:
PV = nRT
Where:
P = Pressure (constant in this case due to isobaric conditions)
V = Volume (the quantity we are trying to find)
n = Number of moles of gas
R = Ideal gas constant
T = Temperature in Kelvin
Step 1: Convert the temperatures to Kelvin
Given:
Initial temperature (T1) = 25˚C
Final temperature (T2) = -196˚C
To convert from Celsius to Kelvin, we use the formula:
Kelvin = Celsius + 273.15
T1 = 25 + 273.15 = 298.15 K
T2 = -196 + 273.15 = 77.15 K
Step 2: Calculate the initial and final volumes
Given:
Initial volume (V1) = 15 cm3
We want to find the final volume (V2).
Step 3: Simplify and rearrange the ideal gas law equation
Since the pressure (P) is constant, it cancels out from both sides of the equation, giving us:
V1/T1 = V2/T2
Step 4: Substitute the known values into the equation
V1 = 15 cm3
T1 = 298.15 K
T2 = 77.15 K
(15 cm3) / (298.15 K) = V2 / (77.15 K)
Step 5: Solve for V2
Cross-multiply and solve:
V2 = (15 cm3) * (77.15 K) / (298.15 K)
V2 ≈ 3.870 cm3
The volume of the helium at 25˚C is approximately 3.870 cm3.