Question

A weather balloon is inflated to a volume of 29.8 L at a pressure of 734 mmHg and a temperature of 27.9 ∘C . The balloon rises in the atmosphere to an altitude where the pressure is 375 mmHg and the temperature is -16.8 ∘C . Assuming the balloon can freely expand, calculate the volume of the balloon at this altitude.

Answer 1

Answer: 49.67 Liters

Explanation: To solve this problem, we only have to use 1 equation: P1V1/T1=P2V2/T2. P, T, and V represent pressure, temperature, and volume respectively. This equation can be used when dealing with gas problems. Usually, there is some change in pressure, temperature, or volume, that will alter pressure or volume. Temperature does not change with pressure or volume.

We are given the balloon starting at 29.8 L(V1), a pressure of 734 mmHg(P1), and a temperature of 27.9 C(T1). The ballon rises and encounters a new set of conditions: 375 mmHg(P2) and -16.8 C(T2). The only variable we are missing is V2 which we will solve for. But before we plug in, we have to do some unit conversions. When using any ideal gas equation, the temperature has to be in Kelvin, the volume in Liters, and the pressure in atm. To convert from mmHg to atm, we simply divide by 760. To convert from Celcius to Kelvin we add 273. Now we can plug into the equation.

((.9657 atm)(29.8 L))/300.8 K=((.4934 atm)(V2))/256.2 K

((.9657 atm)(29.8 L)(256.2 K))/((300.8 K)(.4934 atm))=V2

V2= 49.67 Liters

Hope this Helps!