5. A tank of water vapor at 2.Oatm is compressed until it reaches 10.5atm. If the original temperature
of the water vapor was 115°C, what is the new temperature?
a. Identify the law to be used:
b. Solve the problem:
of the water vapor was 115°C, what is the new temperature?
a. Identify the law to be used:
b. Solve the problem:
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Gay Lussac's Law
P1/T1 = P2/T2
where in, P1 = first pressure
T1 = first temperature
P2 = second pressure
T1 = second temperature
Given: P1 = 2.0atm
T1 = 388.15K
P2 = 10.5atm
Note: Dealing with Temperature in each from the Gas Law must always be in Kelvin.
°C → K
115°C + 273.15K = 388.15K
Unknown: T2 = ?
Formula: T2 = P2T1 / P1
Note: To get the formula, apply cross-multiplication, division and cancellation of like terms using the original formula of Gay Lussac's Law.
P1/T1 = P2/T2 ← cross-multiply
P1T2 = P2T1 ← divide both sides by P1
P1T2 / P1 = P2T1 / P1 ← cancel like terms
T2 = P2T1 / P1 ← final formula for T2
Solution: T2 = P2T1 / P1
= (10.5atm)(388.15K) / 2.0atm
= 4075.58atm•K / 2.0atm
= 2037.79K
Note: "atm" is cancelled out, so the unit for temperature will remain which is "K"
Reminder: Cancellation of units is applicable when the other same unit is on the denominator.
Final Answer: T2 = 2037.79K or 2038K