1) A sample of He has a volume of 0.180 L, a pressure 0.800 atm and a temperatur
ID: 1048330 • Letter: 1
Question
1) A sample of He has a volume of 0.180 L, a pressure 0.800 atm and a temperature of 302 K. At what temperature (K) will the He have a volume of 90.0 mL and a pressure of 3.20 atm. (Hint: Combined Gas Law)Reasoning:
Solution:
2) A cylinder contains 5.0 L of oxygen gas at 20.0 OC and 0.850 ATM. How many grams of oxygen are in the cylinder? (Hint: Ideal Gas Law)
Reasoning:
Solution:
Please show all work! Please and thank you!
1) A sample of He has a volume of 0.180 L, a pressure 0.800 atm and a temperature of 302 K. At what temperature (K) will the He have a volume of 90.0 mL and a pressure of 3.20 atm. (Hint: Combined Gas Law)
Reasoning:
Solution:
2) A cylinder contains 5.0 L of oxygen gas at 20.0 OC and 0.850 ATM. How many grams of oxygen are in the cylinder? (Hint: Ideal Gas Law)
Reasoning:
Solution:
Please show all work! Please and thank you!
1) A sample of He has a volume of 0.180 L, a pressure 0.800 atm and a temperature of 302 K. At what temperature (K) will the He have a volume of 90.0 mL and a pressure of 3.20 atm. (Hint: Combined Gas Law)
Reasoning:
Solution:
2) A cylinder contains 5.0 L of oxygen gas at 20.0 OC and 0.850 ATM. How many grams of oxygen are in the cylinder? (Hint: Ideal Gas Law)
Reasoning:
Solution:
Please show all work! Please and thank you!
Reasoning:
Solution:
2) A cylinder contains 5.0 L of oxygen gas at 20.0 OC and 0.850 ATM. How many grams of oxygen are in the cylinder? (Hint: Ideal Gas Law)
Reasoning:
Solution:
Please show all work! Please and thank you!
Explanation / Answer
(1) We know that PV = nRT
Where
T = Temperature ;P = pressure ; n = No . of moles ;R = gas constant ; V= Volume of the gas
As the gas remains the same , n will constant
So PV / T = constant
Thereby PV/T = P'V'/T'
Where
P = initial pressure = 0.8 atm
V = initial volume = 0.180 L
T = initial temperature = 302 K
P'= final pressure = 3.20 atm
V' = final volume = 90.0 mL = 0.09 L
T' = final temperature = ?
Plug the values we get T' = (P'V'T) / (PV) = 604 K
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(2)
We know that ideal gas equation is PV = nRT
Where
T = Temperature = 20 oC = 20 + 273 = 293 K
P = pressure = 0.850 atm
n = No . of moles = ?
R = gas constant = 0.0821 L atm / mol - K
V= Volume of the gas = 5.0 L
PLug the values we get n = (PV) / (RT)
= 0.177 moles
So mass of oxygen , m = number of moles x molar mass
= 0.177 mol x 32 g/mol
= 5.65 g
Therefore the mass of oxygen in the cylinder is 5.65 g
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