A compartment measures 3.5 meters by 5.5 meters and is 3 meters high. A fire raises the temperature from 68of to 1500 of. If the starting pressure was 1 atmosphere, what volume is present at the elevated temperature assuming the compartment has openings to reduce pressure?
Assuming the compartment has not changed shape during the fire...the volume is the same, whether or not the pressure changes.
3.5*5.5*3=57.75m^2
trick question methinks
m^3
To find the volume at the elevated temperature, we can use the ideal gas law equation, which states:
PV = nRT
Where:
P is the pressure
V is the volume
n is the number of moles of gas
R is the ideal gas constant (0.0821 atm·L/mol·K)
T is the temperature in Kelvin
First, we need to convert the temperatures from Fahrenheit to Kelvin. We can use the following conversion formula:
T(K) = (T(°F) - 32) × 5/9 + 273.15
Initial temperature: T1 = (68 - 32) × 5/9 + 273.15
Final temperature: T2 = (1500 - 32) × 5/9 + 273.15
Next, we can calculate the initial and final volumes using the ideal gas law equation.
For the initial volume:
P1V1 = nRT1
Assuming 1 mole of gas:
P1V1 = (1)(0.0821)(T1)
Now, rearranging the equation to solve for V1:
V1 = (0.0821)(T1) / P1
Substituting the given values:
V1 = (0.0821)(T1) / (1)
Similarly, for the final volume:
P2V2 = nRT2
Again, assuming 1 mole of gas:
P2V2 = (1)(0.0821)(T2)
Now, rearranging the equation to solve for V2:
V2 = (0.0821)(T2) / P2
Substituting the given values:
V2 = (0.0821)(T2) / (1)
Finally, we can calculate the volume at the elevated temperature assuming the compartment has openings that allow pressure to decrease:
V2/V1 = (0.0821)(T2) / P2 / [(0.0821)(T1) / P1]
Simplifying:
V2/V1 = (T2 / T1) * (P1 / P2)
Substituting the given values:
V2/V1 = (T2 / T1) * (1 / 1)
Therefore, the volume at the elevated temperature assuming the compartment has openings to reduce pressure is:
V2 = (T2 / T1) * V1