QuestionSBA
QuestionChoice
PV=nRT
Therefore, n=(PV)/(RT)
The total pressure in the alveoli at the end of inspiration is 100,000Pa. Pressure in the lungs exerted by inhaled air is equal to the total pressure minus the pressure of water added upon hydration: 100,000Pa - 6000Pa = 94,000Pa.
The pressure exerted by perfect gas A in the alveoli is therefore 30% of 94,000Pa = 28,200Pa = 28,200N/m(squared).
Volume of perfect gas A is 30% of 500ml = 150ml = 150cm(cubed) = 150/(100 x 100 x 100)m(cubed) = 0.00015m(cubed).
The temperature in the alveoli in Kelvin is 273 + 37 = 310K.
n=(28,200N/m(squared) x 0.00015m(cubed))/(8.314J/mol.K x 310K)
n=(4.23Nm)/(2521.54J/mol)=(4.23J)/(2521.54J/mol)
n=0.00167754626mol
n=1.6mmol
1.7mmol could have been reached if the temperature was not corrected from 293K to 310K.
10001203
c
1.7mmol
QuestionChoice
PV=nRT
Therefore, n=(PV)/(RT)
The total pressure in the alveoli at the end of inspiration is 100,000Pa. Pressure in the lungs exerted by inhaled air is equal to the total pressure minus the pressure of water added upon hydration: 100,000Pa - 6000Pa = 94,000Pa.
The pressure exerted by perfect gas A in the alveoli is therefore 30% of 94,000Pa = 28,200Pa = 28,200N/m(squared).
Volume of perfect gas A is 30% of 500ml = 150ml = 150cm(cubed) = 150/(100 x 100 x 100)m(cubed) = 0.00015m(cubed).
The temperature in the alveoli in Kelvin is 273 + 37 = 310K.
n=(28,200N/m(squared) x 0.00015m(cubed))/(8.314J/mol.K x 310K)
n=(4.23Nm)/(2521.54J/mol)=(4.23J)/(2521.54J/mol)
n=0.00167754626mol
n=1.6mmol
1.8mmol could have been reached if the water vapour saturated pressure was not corrected for.
10001202
b
1.8mmol
QuestionChoice
PV=nRT
Therefore, n=(PV)/(RT)
The total pressure in the alveoli at the end of inspiration is 100,000Pa. Pressure in the lungs exerted by inhaled air is equal to the total pressure minus the pressure of water added upon hydration: 100,000Pa - 6000Pa = 94,000Pa.
The pressure exerted by perfect gas A in the alveoli is therefore 30% of 94,000Pa = 28,200Pa = 28,200N/m(squared).
Volume of perfect gas A is 30% of 500ml = 150ml = 150cm(cubed) = 150/(100 x 100 x 100)m(cubed) = 0.00015m(cubed).
The temperature in the alveoli in Kelvin is 273 + 37 = 310K.
n=(28,200N/m(squared) x 0.00015m(cubed))/(8.314J/mol.K x 310K)
n=(4.23Nm)/(2521.54J/mol)=(4.23J)/(2521.54J/mol)
n=0.00167754626mol
n=1.6mmol
10001201
a
1.6mmol
QuestionChoice
PV=nRT
Therefore, n=(PV)/(RT)
The total pressure in the alveoli at the end of inspiration is 100,000Pa. Pressure in the lungs exerted by inhaled air is equal to the total pressure minus the pressure of water added upon hydration: 100,000Pa - 6000Pa = 94,000Pa.
The pressure exerted by perfect gas A in the alveoli is therefore 30% of 94,000Pa = 28,200Pa = 28,200N/m(squared).
Volume of perfect gas A is 30% of 500ml = 150ml = 150cm(cubed) = 150/(100 x 100 x 100)m(cubed) = 0.00015m(cubed).
The temperature in the alveoli in Kelvin is 273 + 37 = 310K.
n=(28,200N/m(squared) x 0.00015m(cubed))/(8.314J/mol.K x 310K)
n=(4.23Nm)/(2521.54J/mol)=(4.23J)/(2521.54J/mol)
n=0.00167754626mol
n=1.6mmol
42mol.mmHg.ml/J could have been reached if the standard unit conversions for pressure and volume were not used, instead using mmHg and ml.

<br>
10001205
e
42mmol
QuestionChoice
PV=nRT
Therefore, n=(PV)/(RT)
The total pressure in the alveoli at the end of inspiration is 100,000Pa. Pressure in the lungs exerted by inhaled air is equal to the total pressure minus the pressure of water added upon hydration: 100,000Pa - 6000Pa = 94,000Pa.
The pressure exerted by perfect gas A in the alveoli is therefore 30% of 94,000Pa = 28,200Pa = 28,200N/m(squared).
Volume of perfect gas A is 30% of 500ml = 150ml = 150cm(cubed) = 150/(100 x 100 x 100)m(cubed) = 0.00015m(cubed).
The temperature in the alveoli in Kelvin is 273 + 37 = 310K.
n=(28,200N/m(squared) x 0.00015m(cubed))/(8.314J/mol.K x 310K)
n=(4.23Nm)/(2521.54J/mol)=(4.23J)/(2521.54J/mol)
n=0.00167754626mol
n=1.6mmol
18.6mmol could have been reached if the pressure and volume were not corrected to reflect perfect gas A accounting for 30% rather than 100% of the room air pressure and volume.
10001204
d
18.6mmol
Concept
Chapter
2693
3893
Volume, pressure and compliance
Topic
150
Respiratory physiology
12787
Assume room temperature is 20 degrees Celsius, room pressure is 760mmHg (100,000Pa), room air is dry, saturated air at 37 Celsius contains 47mmHg (6000Pa) of water vapour, the gas constant (R) is 8.314J/mol.K, tidal volume subtract dead space is 500ml, alveoli are water vapour saturated at body temperature and perfect gas A makes up 30% of room air pressure and volume.

How many moles of perfect gas A will be added to the alveoli after a single inhalation?
a