One mole of oxygen gas is at a pressure of 5.10 atm and a temperature of 32.0°C.
(a) If the gas is heated at constant volume until the pressure triples, what is the final temperature?
P V = n R T
V, n and R are constant in this case.
If the pressure P triples at constant volume, so does the absolute termperature, T. It goes from 305 K to 915 K. Convert that back to C, if you wish.
To find the final temperature of the gas when the pressure triples, we can use the ideal gas law. The ideal gas law is given by the equation:
PV = nRT
Where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant
T is the temperature of the gas
In this case, we are given:
Initial pressure (P1) = 5.10 atm
Final pressure (P2) = 5.10 atm x 3 = 15.30 atm (since the pressure triples)
Initial temperature (T1) = 32.0°C
Number of moles of gas (n) = 1 mole (since we have one mole of oxygen gas)
First, we need to convert the initial temperature from Celsius to Kelvin, as the ideal gas law requires temperature in Kelvin.
T1 (Kelvin) = T1 (Celsius) + 273.15
T1 (Kelvin) = 32.0 + 273.15 = 305.15 K
Now, we can rearrange the ideal gas law equation to solve for the final temperature (T2):
P1V1/T1 = P2V2/T2
Since the volume is constant in this case (constant volume), we can cancel out the V1 and V2 terms:
P1/T1 = P2/T2
Now, we can substitute the given values into the equation and solve for T2:
(5.10 atm) / (305.15 K) = (15.30 atm) / T2
Cross-multiplying:
5.10 atm * T2 = 15.30 atm * 305.15 K
Dividing both sides by 5.10 atm:
T2 = (15.30 atm * 305.15 K) / 5.10 atm
T2 ≈ 914.85 K
Finally, let's convert the final temperature from Kelvin back to Celsius:
T2 (Celsius) = T2 (Kelvin) - 273.15
T2 (Celsius) ≈ 914.85 - 273.15 ≈ 641.70°C
Therefore, the final temperature of the gas when the pressure triples is approximately 641.70°C.