Consider 1.5 L of air in a patents lungs at 37.0 C and 1.00 atm pressure. What is the volume would this air occupy if it were at 25 C under a pressure of 5.00 x 10^2 atm?

0.00288 L = 2.88 mL.

Explanation:

To calculate the no. of moles of a gas, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant.

T is the temperature of the gas in K.

If n is constant, and have two different values of (P, V and T):

P₁V₁T₂ = P₂V₂T₁

P₁ = 1.0 atm, V₁ = 1.5 L, T₁ = 37°C + 273 = 310 K.

P₂ = 500.0 atm, V₂ = ? L, T₂ = 25°C + 273 = 298 K.

∴ V₂ = P₁V₁T₂/P₂T₁ = (1.0 atm) (1.5 L) (298 K) / (500.0 atm) (310 L) = 0.00288 L = 2.88 mL.