A weather balloon is inflated to a volume of 25.6L at a pressure of 736mmHg and a temperature of 31.9∘C. The balloon rises in the atmosphere to an altitude, where the pressure is 365mmHg and the temperature is -15.6∘C.

Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

The answer I got is 61.1 L but this is not correct. I do not understand what I am doing wrong.

Why didn't you show your work. Then we could find the trouble in an instant.

Use (P1V1/T1) = (P2V2/T2) and remember T must be in kelvin.

That is the equation that I used.

P1V1T2/T1P2 = (736mmHg x 25.6 L x 304.9 K)/(365mmHg x 257.4K) = 61.1 L

You forgot to change the units of P from mmHg to atm. To do this, just divide the mmHg value by 760 mmHg/atm.

I've tried it by changing mmHg to atm and I still get 61.1L

To calculate the volume of the balloon at the higher altitude, we can use the ideal gas law, which states:

PV = nRT

Where:
P = Pressure
V = Volume
n = number of moles (constant)
R = Ideal gas constant (constant)
T = Temperature

We can rearrange this equation to solve for V:

V = (nRT) / P

However, since we are assuming the balloon freely expands, the number of moles (n) remains constant. Therefore, we can simplify the equation:

V₁ / P₁ = V₂ / P₂

Where:
V₁ = Initial volume
P₁ = Initial pressure
V₂ = Final volume (which we wish to calculate)
P₂ = Final pressure

We can plug in the values given in the problem:

V₁ = 25.6 L
P₁ = 736 mmHg
P₂ = 365 mmHg

Now we can solve for V₂:

V₂ = (V₁ * P₂) / P₁

Substituting the values:

V₂ = (25.6 L * 365 mmHg) / 736 mmHg

Calculating this:

V₂ = 12.71 L

Therefore, at the higher altitude, the volume of the balloon is approximately 12.71 liters.