Under water where the temperature is 17C and the pressure is 394 kPa, a diver inhales 2.1 L of air from his SCUBA tank.
a.How many moles of gas are in his lungs?
b.If the diver swims to the surface without exhaling where the temperature is 32C and the pressure changes to 100.2 kPa, what will the volume of the air in his lungs be?
a. To find the number of moles of gas in the diver's lungs, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure (in kPa)
V = volume (in liters)
n = number of moles
R = ideal gas constant (0.0821 L⋅atm/(mol⋅K) or 8.314 J/(mol⋅K))
T = temperature (in Kelvin)
First, let's convert the temperature from Celsius to Kelvin:
17°C + 273.15 = 290.15 K
Now we can calculate the number of moles of gas:
PV = nRT
(394 kPa)(2.1 L) = n(8.314 J/(mol⋅K))(290.15 K)
Simplifying the equation:
820.74 = 2406.5091n
Solving for n (number of moles):
n = 820.74 / 2406.5091
n ≈ 0.341 mol
Therefore, there are approximately 0.341 moles of gas in the diver's lungs.
b. To find the new volume of air in the lungs as the diver swims to the surface, we can use the combined gas law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Where:
P1 = initial pressure (394 kPa)
V1 = initial volume (2.1 L)
T1 = initial temperature (290.15 K)
P2 = final pressure (100.2 kPa)
V2 = final volume (unknown)
T2 = final temperature (32°C + 273.15 = 305.15 K)
Simplifying the equation:
(P1 * V1) / T1 = (P2 * V2) / T2
Rearranging the equation and solving for V2 (final volume):
V2 = (P2 * V1 * T2) / (P1 * T1)
V2 = (100.2 kPa * 2.1 L * 305.15 K) / (394 kPa * 290.15 K)
Converting units:
V2 = (100.2 kPa * 2.1 L * 305.15 K) / (394 kPa * 290.15 K)
V2 ≈ 1.89 L
Therefore, the volume of air in the diver's lungs will be approximately 1.89 L when he reaches the surface without exhaling.
To determine the number of moles of gas in the diver's lungs, we can use the Ideal Gas Law equation:
PV = nRT
where:
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of gas
R is the ideal gas constant (0.0821 L·atm/mol·K)
T is the temperature in Kelvin
a. First, we need to convert the given temperature from Celsius to Kelvin:
T1 = 17°C + 273.15 = 290.15 K
Next, we can substitute the given values into the equation to solve for n:
394 kPa * 2.1 L = n * 0.0821 L·atm/mol·K * 290.15 K
Simplifying the equation:
n = (394 kPa * 2.1 L) / (0.0821 L·atm/mol·K * 290.15 K)
n ≈ 21.4 moles
Therefore, there are approximately 21.4 moles of gas in the diver's lungs.
b. To find the new volume of air in the lungs when the diver swims to the surface, we can use the Combined Gas Law equation:
(P1 * V1) / T1 = (P2 * V2) / T2
where:
P1 is the initial pressure of the gas
V1 is the initial volume of the gas
T1 is the initial temperature of the gas in Kelvin
P2 is the final pressure of the gas
V2 is the final volume of the gas (what we are trying to find)
T2 is the final temperature of the gas in Kelvin
First, we convert the given temperatures to Kelvin:
T1 = 17°C + 273.15 = 290.15 K
T2 = 32°C + 273.15 = 305.15 K
Next, let's substitute the given and calculated values into the equation to solve for V2:
(394 kPa * 2.1 L) / 290.15 K = (100.2 kPa * V2) / 305.15 K
Simplifying the equation:
V2 = (394 kPa * 2.1 L * 305.15 K) / (290.15 K * 100.2 kPa)
V2 ≈ 4.36 L
Therefore, when the diver swims to the surface without exhaling, the volume of air in his lungs will be approximately 4.36 L.
For a. Use PV = nRT. You know P and V, R and T. Solve for n.
For b, using n from part a, then use the new P and T to solve for V.
By the way, you can use kPa directly, if you wish, without changing to atmospheres IF you use 8.3145 for R.