A 30.0 L sample of nitrogen inside a rigid, metal container at 20.0 degrees C is placed inside an oven whose temperature is 50.0 degrees D. The pressure inside the container at 20.0 degrees C was at 3.00 atm. What is the pressure of the nitrogen after its temperature is increased?
Pf = 3.31 atm
To find the pressure of the nitrogen after its temperature is increased, we can use the combined gas law equation:
(P1 * V1) / (T1) = (P2 * V2) / (T2)
Given:
P1 = 3.00 atm (initial pressure)
V1 = 30.0 L (initial volume)
T1 = 20.0 °C = 20.0 + 273.15 = 293.15 K (initial temperature)
We need to find P2 (final pressure) when T2 (final temperature) is given as 50.0 °D. However, we need to convert the temperature to Kelvin since the gas law equation requires temperature in Kelvin.
Now, let's convert 50.0 °D to Kelvin:
T2 in Kelvin = (50 - 32) × (5/9) + 273.15
T2 in Kelvin ≈ 283.15 K
Substituting the given values into the combined gas law equation:
(3.00 atm * 30.0 L) / (293.15 K) = (P2 * 30.0 L) / (283.15 K)
To find P2 (final pressure), we can rearrange the equation as follows:
P2 = (3.00 atm * 30.0 L * 283.15 K) / (293.15 K * 30.0 L)
P2 ≈ 3.31 atm
Therefore, the pressure of the nitrogen after its temperature is increased is approximately 3.31 atm.