(For first 4 questions) A sample of argon gas has a volume of 755 mL at a pressu
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Question
(For first 4 questions) A sample of argon gas has a volume of 755 mL at a pressure of 1.32 atm and a temperature of 130 C.
(1) What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 658 mmHg and 305 K , if the amount of gas remains the same?
(2) What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 0.750 atm and 82 C, if the amount of gas remains the same?
(3) What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 0.750 atm and 82 C, if the amount of gas remains the same?
(4) What is the volume of the gas in milliliters when the pressure and temperature of the gas sample are changed to 16.4 atm and -17 C, if the amount of gas remains the same?
(For Next 3 Questions) A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature.
(1) A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature. What is the new volume, in liters, after 0.600 mole of O2 gas is added to the initial sample 6.40 g of O2?
(2) A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature. If oxygen is released until the volume is 10.0 L how many moles of O2 are removed?
(3)A sample containing 6.40 g of O2 gas has a volume of 27.0 L at constant pressure and temperature. What is the volume, in liters, after 6.00 g of He is added to the O2 gas in the container?
Explanation / Answer
V1 = 755 mL ,
P1 = 1.32 atm ,
T1 = 130 + 273 = 403 K
1)
P2 = 658 mmHg = 0.866 atm
T2 = 305 K
V2 = ?
P1 V1 / T1 = P2 V2 / T2
1.32 x 755 / 403 = 0.866 x V2 / 305
V2 = 871 mL
volume = 871 mL
2)
P2 = 0.750 atm
T2 = 82 +273 = 355 K
V2 = ?
P1 V1 / T1 = P2 V2 / T2
1.32 x 755 / 403 = 0.750x V2 / 355
V2 = 1170 mL
volume = 1170 mL
3)
P2 = 0.750 atm
T2 = 82 +273 = 355 K
V2 = ?
P1 V1 / T1 = P2 V2 / T2
1.32 x 755 / 403 = 0.750x V2 / 355
V2 = 1170 mL
volume = 1170 mL
4)
P2 = 16.4 atm
T2 = -17 +273 = 256 K
V2 = ?
P1 V1 / T1 = P2 V2 / T2
1.32 x 755 / 403 = 16.4 x V2 / 256
V2 = 0.0386 mL
volume = 0.0386 mL
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