Question

A person takes in a breath of −9 ◦C air and holds it until it warms to 37.3 ◦C. The air has an initial volume of 0.72 L and a mass of 0.00077 kg.

a) Determine the work done by the air on the lungs if the pressure remains constant at 1 atm. Answer in units of J.

b) Find the change in the internal energy of the air. Treat the air as a monatomic gas. Answer in units of J.

c) Calculate the energy added to the air by heat. Answer in units of J.

Answer 1

initial temperature T1 = -9 + 273 = 264 K

final temperature T2 = 37.3+273 = 310.3 K

initial volume V1 = 0.72 L = 0.72*10^-3 m^3

initial pressure P1 = 1 atm = 10^5 Pa

a)

the pressure is constant

T2/T1 = V2/V1

V2 = (T2/T1)*V1

V2 = (310.3/264)*0.72*10^-3 = 0.846*10^-3 m^3

at constant pressure

work done W = P*dV = P*(V2-V1)

W = 10^5*(0.846*10^-3-0.72*10^-3)

W = 12.6 J

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(b)

change in internal energy dU = n*Cv*dT = n*(3/2)*R*dT = (3/2)*nR*(T2-T1) = (3/2)*(P*V2 - P*V1)

change in internal energy dU = (3/2)*P*(V2-V1) = (3/2)*10^5*(0.846*10^-3-0.72*10^-3) = 18.9 J

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(c)

from Ist law of thermodynamics

Q = dU + W

Q = 31.5 J