I am studying Fire Science Engineering. I have a question concerning my homework. The question is A compartment measures 3 meters by 5 meters and is 2.8 meters high. A fire raises the temperature from 20 degrees C to 1000 degrees C. If the starting pressure was one atmosphere, what pressure is present at the elevated temperature assuming the compartment remains closed? This will be appreciated. Thanks.
If the compartment is closed, the number of moles and volume are constant, so the ideal gas law tells you that
P/T = constant
where T is the absolute temperature.
P2/T2 = P1/T1
P2 = (T2/T1)*P1 = 1273/293 = 4.345 atm
Thank you so much for the help. Appreciate it. Take Care.
To solve this question, we can use the ideal gas law equation, which states that the pressure of a gas is proportional to its temperature, volume, and the number of moles present. The ideal gas law equation is:
PV = nRT
Where:
P = pressure of the gas
V = volume of the compartment
n = number of moles of gas
R = ideal gas constant
T = temperature of the gas in Kelvin
At the start, the temperature is 20 degrees C, which we need to convert to Kelvin using the equation:
T(K) = T(C) + 273.15
Given:
Initial temperature (T1) = 20 degrees C
Final temperature (T2) = 1000 degrees C
Initial pressure (P1) = 1 atmosphere
Volume (V) = 3m x 5m x 2.8m
First, we convert the temperatures to Kelvin:
T1(K) = 20 + 273.15 = 293.15 K
T2(K) = 1000 + 273.15 = 1273.15 K
Next, we calculate the ratio of the pressures using the ideal gas law equation:
P1V1 / nR = P2V2 / nR
Since the number of moles and the ideal gas constant are the same on both sides, we can simplify the equation to:
P1V1 = P2V2
Now, we substitute the given values into the equation:
P1 = 1 atmosphere
V1 = 3m x 5m x 2.8m = 42m³
V2 = 42m³ (volume remains constant as the compartment is closed)
P1V1 = P2V2
1atm * 42m³ = P2 * 42m³
Simplifying the equation further:
42 = P2
Therefore, the pressure at the elevated temperature of 1000 degrees Celsius is 42 atmospheres (assuming the compartment remains closed).