. An open tank holding 400 cubic foot of air at atmospheric pressure at 70 degrees F and heated to 160 degree F. What is the volume of air that escapes from the tank?
To calculate the volume of air that escapes from the tank when heated, we need to use the ideal gas law equation, which states that the product of pressure (P), volume (V), and temperature (T) is proportional to the number of molecules of gas (n) and the gas constant (R).
The equation is as follows:
PV = nRT
Where:
P = Pressure
V = Volume
n = Number of molecules
R = Gas constant
T = Temperature in Kelvin
To begin, let's convert the temperatures to Kelvin. The conversion from Fahrenheit to Kelvin is done by adding 459.67 to the Fahrenheit temperature.
Given that the initial temperature is 70 degrees F and the final temperature is 160 degrees F, we can convert them to Kelvin as follows:
Initial temperature (T1) = 70 + 459.67 = 529.67 K
Final temperature (T2) = 160 + 459.67 = 619.67 K
Now, let's substitute the known values into the ideal gas law equation:
P1V1 = nRT1 (Initial state of the gas)
P2V2 = nRT2 (Final state of the gas)
For both equations, the pressure (P) and the number of molecules (n) remain constant since the question states that the tank is at atmospheric pressure. Therefore, we can conclude that P1 = P2 and n1 = n2.
Since these values are the same, we can simplify the equations to:
V1 = RT1 (Initial volume)
V2 = RT2 (Final volume)
To find the volume of air that escaped from the tank, we need to calculate the difference between the final volume (V2) and the initial volume (V1).
ΔV = V2 - V1
Now, let's plug in the known values. The gas constant (R) is approximately 0.73 ft³·psi/(lb·R) for air.
V1 = RT1 = 0.73 × 529.67 ft³·psi/(lb·R) = 386.8749 ft³
V2 = RT2 = 0.73 × 619.67 ft³·psi/(lb·R) = 452.5331 ft³
ΔV = V2 - V1 = 452.5331 ft³ - 386.8749 ft³ = 65.6582 ft³
Therefore, approximately 65.6582 cubic feet of air escape from the tank when heated from 70 degrees F to 160 degrees F.