Question

How many moles of gas would it take to fill an average man's lungs, total capacity of which is about 4.5 L. Assume 1.00 atm pressure and 37 degrees C

Answer 1

0.124 moles

Step-by-step explanation:

PV = nRT, can be used, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature in Kelvin.

Converting the volume to liters and temperature to Kelvin:

V = 45 L

T = 37°C + 273.15 = 310.15 K

Plugging these values into the ideal gas law:

1.00 atm * 45 L = n * 0.0821 Latm/mol/K * 310.15 K

Solving for n:

n = (1.00 atm * 45 L) / (0.0821 Latm/mol/K * 310.15 K) = 0.124 mol

So, it would take about 0.124 moles of gas to fill an average man's lungs.

Answer 2

0.0144 moles

Step-by-step explanation:

To calculate the number of moles of gas in a volume of 4.5 liters, we use the ideal gas law, PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

First, we need to convert the temperature to Kelvin:

37°C = 37 + 273.15 = 310.15 K

Next, we can plug in the known values into the ideal gas law and solve for n:

PV = nRT

(1.00 atm)(4.5 L) = (n)(0.0821 Latm/mol K)(310.15 K)

n = (1.00 atm)(4.5 L) / [(0.0821 Latm/mol K)(310.15 K)] = 0.0144 moles

So, it would take approximately 0.0144 moles of gas to fill an average man's lungs.


ALLLEN