Two identical ballons are filled, one w/ argon the other w/ oxygen gas at the same temp. If the oxygen ballon leaks at the rate of 64 mL per hour, how many hours will it take the argon ballon to lose 195mL of argon gas?
formula:
rate 1/rate 2=square root of m1/m2
You have posted the wrong formula.
Its r1/r2 = sqrt(M2/M1).
Let me know if that doesn't do it for you.
To find the number of hours it will take for the argon balloon to lose 195 mL of argon gas, we need to use the given formula:
rate1 / rate2 = √(m1 / m2)
Where:
rate1 is the leak rate of the oxygen balloon (64 mL/hour)
rate2 is the leak rate of the argon balloon (unknown)
m1 is the initial amount of gas in the oxygen balloon (unknown)
m2 is the initial amount of gas in the argon balloon (unknown)
We can rearrange the formula to solve for rate2:
rate2 = rate1 * √(m1 / m2)
Since the balloons are identical and filled at the same temperature, we can assume that m1 = m2.
Let's substitute the given values into the formula and solve for rate2:
rate2 = 64 * √(m1 / m1)
Since m1 = m2, the formula simplifies to:
rate2 = 64 * √(1)
The square root of 1 is 1, so the formula further simplifies to:
rate2 = 64 * 1
rate2 = 64 mL/hour
Therefore, the leak rate of the argon balloon is also 64 mL/hour.
Now, to find the number of hours it will take for the argon balloon to lose 195 mL of argon gas, we can set up the following equation:
rate2 * time = 195
Substituting rate2 with 64 mL/hour:
64 * time = 195
Solving for time:
time = 195 / 64
time ≈ 3.04 hours
Therefore, it will take approximately 3.04 hours for the argon balloon to lose 195 mL of argon gas.