A scuba diver takes a tank of air on a deep dive. The tank's volume is 10 L and it is completely filled with air at an absolute pressure of 224 atm at the start of the dive. The air temperature at the surface is 92 F and the diver ends up in deep water at 50 F. Assuming thermal equilibrium and neglecting air loss, determine the absolute internal pressure of the air when it is cold.
P/T is a constant at constant (tank) volume and amount (moles) of gas. T must be an absolute temperature. Either Kelvin or Rankine will work.
T1 = 92F = 552 R
T2 = 50F = 510 R
"R" is the absolute Rankine temperature. You get it by adding 460 to the Fahrenheit temperature.
P2/P1 = T2/T1 = 510/552 = 0.9239
P2 = 207 atm
To determine the absolute internal pressure of the air when it is cold, we need to use the ideal gas law equation:
PV = nRT
Where:
P = absolute pressure of the gas
V = volume of the gas
n = number of moles of the gas
R = ideal gas constant
T = temperature of the gas in Kelvin
First, let's convert the temperatures to Kelvin. We know that 0°C is equal to 273.15 K, so:
Temperature at the surface = 92°F = (92 - 32) × (5/9) + 273.15 = 305.37 K
Temperature in deep water = 50°F = (50 - 32) × (5/9) + 273.15 = 283.15 K
Next, we need to determine the number of moles of air in the tank. To do this, we need to use the ideal gas law again, but isolate the number of moles (n) this time:
n = (PV) / (RT)
Given:
P = 224 atm
V = 10 L
R = 0.0821 L·atm/(mol·K)
T = 305.37 K (temperature at the surface)
n = (224 atm × 10 L) / (0.0821 L·atm/(mol·K) × 305.37 K)
n ≈ 92.33 mol
Now, using the above value of n, we can calculate the absolute pressure at the lower temperature using the ideal gas law:
P2 = (nRT2) / V
Given:
n = 92.33 mol (number of moles of air)
R = 0.0821 L·atm/(mol·K)
T2 = 283.15 K (temperature in deep water)
V = 10 L
P2 = (92.33 mol × 0.0821 L·atm/(mol·K) × 283.15 K) / 10 L
P2 ≈ 210.02 atm
Therefore, the absolute internal pressure of the air when it is cold is approximately 210.02 atm.