A sample of methane (CH4) has a volume of 26 mL at a pressure of 0.85 atm . What
ID: 956661 • Letter: A
Question
A sample of methane (CH4) has a volume of 26 mL at a pressure of 0.85 atm . What is the volume, in milliliters, of the gas at each of the following pressures if there is no change in temperature and amount of gas? 1. 0.35 atm 2. 2.30 atm 3. 3100 mmHg The volume of air in a person's lungs is 626 mL at a pressure of 761 mmHg . Inhalation occurs as the pressure in the lungs drops to 731 mmHg with no change in temperature and amount of gas. 1. To what volume, in milliliters, did the lungs expand? sample of neon initially has a volume of 3.00 L at 16 C . 1. What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 5.50 L ? 2. What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 1500 mL ? 3. What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 8.50 L ? 4. What is the new temperature, in degrees Celsius, when the volume of the sample is changed at constant pressure and amount of gas to 3700 mL ? A balloon contains 2500 mL of helium gas at 85 C. What is the new volume of the gas when the temperature changes to the following, if n (the amount of gas) and P are not changed? 1. 40 C 2. 660 K 3. -15 C 4. 235 K
Explanation / Answer
1) Data.
P1 = 0.85 atm
V1 = 26 ml
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a.
P2 = 0.35 atm
P1V1 = P2V2
V2 = 0.85atmx26ml/0.35atm = 63.14 ml
b.
P2 = 2.30 atm
P1V1 = P2V2
V2 = 0.85atmx26ml/2.30atm = 9.61 ml
c.
P2 = 3100 mmHg = 4.08 atm
P1V1 = P2V2
V2 = 0.85atmx26ml/4.08atm = 5.42 ml
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2) Data.
P1 = 761 mmHg
V1 = 626 ml
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P2 = 731 mmHg
P1V1 = P2V2
V2 = 761mmHgx626ml/731mmHg = 651.7 ml
Expantion of lungs by Volume = V2 - V1 = 651.7 - 626 = 25.7 ml
The lungs expand by 25.7 ml
3) Data.
Neon Gas (Ne)
V1 = 3.00 L
T1 = 16°C
P1 = P2 = 1 atm (asumption)
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a.
V2 = 5.50 L
P1V1/T1 = P2V2/T2
T2 = P2V2T1 / P1V1
T2 = 1atm x 5.50L x 16°C / 1atm x 3L
T2 = 29.34°C
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b.
V2 = 1500 ml = 1.50 L
P1V1/T1 = P2V2/T2
T2 = P2V2T1 / P1V1
T2 = 1atm x 1.50L x 16°C / 1atm x 3L
T2 = 8°C
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c.
V2 = 8.50 L
P1V1/T1 = P2V2/T2
T2 = P2V2T1 / P1V1
T2 = 1atm x 8.50L x 16°C / 1atm x 3L
T2 = 45.34°C
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d.
V2 = 3700 ml = 3.70 L
P1V1/T1 = P2V2/T2
T2 = P2V2T1 / P1V1
T2 = 1atm x 3.70L x 16°C / 1atm x 3L
T2 = 19.74°C
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4) Data.
V1 = 2500ml
T1 = 85°C
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V1/T1 = V2/T2
V2 = V1xT2/T1
a.
T2 = 1.40°C
V2 = 2500ml x 1.40°C / 85°C
V2 = 41.18 ml
b.
T2 = 660°K
V2 = 2500ml x 660°K / (85+273)°K
V2 = 4609 ml
c.
T2 = -15°C
V2 = 2500ml x (-15°C+273)°K / (85+273)°K
V2 = 1802 ml
d.
T2 = 235°K
V2 = 2500ml x 235°K / (85+273)°K
V2 = 1641 ml
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Cheers!
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