A Balloon with a volume of 2.00 L is at a pressure of 1.5 atm and a temperature of 20C. If the balloon is put into a container and the pressure is increased to 2.0 atm and the temperature is raised to 25C, what is the new volume of the balloon?

To calculate the new volume of the balloon, we can use the ideal gas law equation:

PV = nRT

Where:
P = pressure
V = volume
n = number of moles of gas
R = ideal gas constant
T = temperature

The ideal gas law equation assumes that the number of moles of gas remains constant. Therefore, to find the new volume, we need to consider the initial and final conditions and calculate any changes in pressure and temperature.

Given:
Initial volume (V1) = 2.00 L
Initial pressure (P1) = 1.5 atm
Initial temperature (T1) = 20°C = 20 + 273.15 = 293.15 K

Final pressure (P2) = 2.0 atm
Final temperature (T2) = 25°C = 25 + 273.15 = 298.15 K

Using the ideal gas law equation, we have:

(P1)(V1) = (P2)(V2)

We can rearrange the equation to solve for the new volume (V2):

V2 = (P1)(V1) / P2

Now, let's plug in the values:

V2 = (1.5 atm)(2.00 L) / 2.0 atm
V2 = 1.5 L

Therefore, the new volume of the balloon is 1.5 L.

It appears to me that if you put the balloon into a container that it may well fill the container; therefore, the volume of the balloon will be the size of the container. Also, I wonder if the second pressure is inside the balloon or outside the balloon? And is the container a closed vessel?