Question
The air inside a balloon is at a temperature of 300 K and a pressure of 2.00 atm. If the balloon contains 1.60 mol of air, what is its volume?
R= 0.0821 L atm/mol K
PV = nRT
R= 0.0821 L atm/mol K
PV = nRT
Answers
Answer
a
19.7 L
b
1.7 L
c
7.70 L
d
30.8 L
Answered by
GPT 3.5
To find the volume of the balloon, we can use the ideal gas law equation:
PV = nRT
Where:
P = pressure = 2.00 atm
V = volume (unknown)
n = number of moles = 1.60 mol
R = gas constant = 0.0821 L atm/mol K
T = temperature = 300 K
Plugging in the values:
(2.00 atm)(V) = (1.60 mol)(0.0821 L atm/mol K)(300 K)
Simplifying:
2.00V = 39.312 L atm
Dividing both sides by 2.00:
V = 19.656 L
Therefore, the volume of the balloon is approximately 19.656 L.
PV = nRT
Where:
P = pressure = 2.00 atm
V = volume (unknown)
n = number of moles = 1.60 mol
R = gas constant = 0.0821 L atm/mol K
T = temperature = 300 K
Plugging in the values:
(2.00 atm)(V) = (1.60 mol)(0.0821 L atm/mol K)(300 K)
Simplifying:
2.00V = 39.312 L atm
Dividing both sides by 2.00:
V = 19.656 L
Therefore, the volume of the balloon is approximately 19.656 L.
Answered by
GPT 3.5
The correct answer is a) 19.7 L.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.