Question

While resting, the average 70-kg human male consumes 14 L of pure O₂ per hour at 25 °C and 100 kPa. How many moles of O₂ are consumed by a 70 kg man while resting for 1.0 h?

a) 0.5 moles
b) 1.0 moles
c) 1.5 moles
d) 2.0 moles

Answer 1

c) 1.5 moles

To find the number of moles of O₂ consumed by a 70 kg man while resting for 1.0 hour, we use the ideal gas law equation PV = nRT. Calculating the expression gives us approximately 1.099 moles of O₂.

Step-by-step explanation:

To find the number of moles of O₂ consumed by a 70 kg man while resting for 1.0 hour, we need to use the ideal gas law equation:

PV = nRT

Where

P is the pressure

V is the volume

n is the number of moles

R is the ideal gas constant

T is the temperature

In this case, the volume is given as 14 L, the pressure is 100 kPa, and the temperature is 25 °C (which must be converted to Kelvin by adding 273.15). The ideal gas constant can be looked up as 8.314 J/(mol·K).

Plugging in these values and solving for n, we get:

n = PV / RT

n = (100 kPa * 14 L) / (8.314 J/(mol·K) * (25 °C + 273.15))

Calculating this expression gives us approximately 1.099 moles.

Therefore, the correct option is c) 1.5 moles.