Question

a weather balloon is filled to a volume of 6000 l while it is on the ground, at a pressure of 1 atm and a temperature of 273 k. the balloon is then released into the atmosphere. as the balloon rises, the pressure decreases to 0.3 atm and the temperature decreases to 240 k. what is the final volume of the balloon at this temperature and pressure? round your answer to the nearest tenth. l

Explanation:

Step1: Recall Combined Gas Law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), where \(P_1 = 1\) atm, \(V_1 = 6000\) L, \(T_1 = 273\) K, \(P_2 = 0.3\) atm, \(T_2 = 240\) K, and \(V_2\) is the final volume we need to find.

Step2: Rearrange the formula to solve for \(V_2\)

From \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), we can solve for \(V_2\) by cross - multiplying: \(V_2=\frac{P_1V_1T_2}{P_2T_1}\)

Step3: Substitute the given values into the formula

Substitute \(P_1 = 1\) atm, \(V_1 = 6000\) L, \(T_1 = 273\) K, \(P_2 = 0.3\) atm, and \(T_2 = 240\) K into the formula for \(V_2\):
\(V_2=\frac{1\times6000\times240}{0.3\times273}\)
First, calculate the numerator: \(1\times6000\times240 = 1440000\)
Then, calculate the denominator: \(0.3\times273=81.9\)
Now, divide the numerator by the denominator: \(V_2=\frac{1440000}{81.9}\approx17582.4\)

Answer:

\(17582.4\)