Human lung capacity varies from about 4L to 6L, so we shall use an average of 5L

Question

Human lung capacity varies from about 4L to 6L, so we shall use an average of 5L. The air enters at the ambient temperature of the atmosphere and must be heated to internal body temperature at an approximately constant pressure of 1.00atm in our model. Suppose you are outside on a winter day when the temperature is -5.00F.

How many moles of air does your lung hold if the 5L is at the internal body temperature of 37C?

How much heat must your body have supplied to get the 5L of air up to internal body temperature, assuming that the atmosphere is all N2 ?

Suppose instead that you manage to inhale the full 5L of air in one breath and hold it in your lungs without expanding (or contracting) them. How much heat would your body have had to supply in that case to raise the air up to internal body temperature?

Explanation / Answer

Volume = 5.0 L temperature = 37 C = 310 K P = 1 atm . Moles of air, n = P V / RT                        =1 atm * 5.0 L / (0.0821 L-atm/ (mol.K) * 310 K)                         =0.196 moles . Temperature outside = -14 F = -25.6 C Temperature required = 37 C Temperature difference = 37 - (-25.6) = 62.6 C . Air is assumed to be completely nitrogen. Specific heat of nitrogen = 1.04 J/ (g.K) mass of nitrogen (air) = moles * molar mass                                  =0.196 moles * 28.00 g/mol                                  =5.5 g . Heat required = mass * specific heat * temperaturedifference                      = 5.5 g * 1.04 J/ (g.K) * 62.6 C                      = 358.1 Joules Volume = 5.0 L temperature = 37 C = 310 K P = 1 atm . Moles of air, n = P V / RT                        =1 atm * 5.0 L / (0.0821 L-atm/ (mol.K) * 310 K)                         =0.196 moles . Temperature outside = -14 F = -25.6 C Temperature required = 37 C Temperature difference = 37 - (-25.6) = 62.6 C . Air is assumed to be completely nitrogen. Specific heat of nitrogen = 1.04 J/ (g.K) mass of nitrogen (air) = moles * molar mass                                  =0.196 moles * 28.00 g/mol                                  =5.5 g . Heat required = mass * specific heat * temperaturedifference                      = 5.5 g * 1.04 J/ (g.K) * 62.6 C                      = 358.1 Joules

Related Questions

Navigate

• Browse (All)
• Browse H
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.