Helium gas in a balloon occupies 2.5 L at 300.0 K. The balloon is dipped into liquid nitrogen that is at a temperature of 80.0 k. What will the volume of the helium in the balloon at the lower temperature be?
I got a weird answer like .66667 and i used charles law
This is not a answer choice for my test. It seems that may be wrong.
thanku!
To solve this problem, you can use Charles's Law, which states that for a fixed amount of gas at a constant pressure, the volume is directly proportional to the temperature.
The equation for Charles's Law is V1/T1 = V2/T2, where V1 and T1 are the initial volume and temperature, and V2 and T2 are the final volume and temperature.
Let's use this equation to solve the problem.
Given:
V1 = 2.5 L (initial volume)
T1 = 300.0 K (initial temperature)
T2 = 80.0 K (final temperature)
Plug these values into the equation:
V1 / T1 = V2 / T2
2.5 L / 300.0 K = V2 / 80.0 K
Solving for V2, we can cross-multiply:
(2.5 L) * (80.0 K) = (300.0 K) * V2
200 L*K = 300.0 K * V2
Divide both sides by 300.0 K:
V2 = (200 L*K) / 300.0 K
V2 ≈ 0.66667 L
Therefore, the volume of the helium in the balloon at the lower temperature (80.0 K) will be approximately 0.66667 L.
It appears that your answer of 0.66667 is correct.
assuming pressure would be the same (it is not if one is submerged in a liquid).
V1/T1=V2/T2
V2=2.5*80/(300)=2.5*4/15 which is about in my head 1/4*2.5 which is your answer