Explanation Given info: The basal metabolic rate is 120 W , the volume of the breath is 0.600 L , the rate of breathe out is 22.0 breaths per minute , the temperature of the air is 37.0 ° C , the vapor pressure is 3.20 kPa , the latent heat of vaporization at 37.0 ° C is same as the heat of vaporization at 100 ° C . Write the expression for the volume of the breathe taken in one second. V breaths = N V Here, N is the number of breaths per second. V is the volume of breaths. Substitute 0.600 L for V and 22.0 breaths per minute for N in above equation. V breaths = ( 0.600 L × 1 m 3 1000 L ) ( 22.0 breaths per minute × 1 s 60 min ) = 2.2 × 10 − 4 m 3 Thus, the volume of the breaths is 2.2 × 10 − 4 m 3 . Write the expression for the ideal gas equation. P V breaths = n R T Here, P is the pressure. T is the temperature. V is the volume. R is the ideal gas constant. n is the number of moles. Substitute 2.2 × 10 − 4 m 3 for V breaths , 3.20 kPa for P , 8.31 J / mol ⋅ K for R and 37.0 ° C for T in above equation. ( 3.20 kPa × 10 3 Pa 1 kPa ) ( 2.2 × 10 − 4 m 3 ) = n ( 8.31 J / mol ⋅ K ) ( 37.0 °C + 273 ) K ≈ 2

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Author: SERWAY

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Chapter 20, Problem 20.70AP

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The rate on which the energy is lost on exhaling humid air.

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