Question

a gas filled balloon with a volume of 12.5 l at 0.90 atm and 294 k is allowed to rise to the stratosphere where the temperature is 268 k and the pressure is 1.0 millibar. what is the final volume of the balloon? (1.0 atm = 1.013 bar). 1.2 l 1.5×10² l 1.0×10⁴ l 1.0×10² l 2.0×10⁴ l

Explanation:

Step1: Convert pressure units

Given \(1.0\ atm = 1.013\ bar\), so \(1\ bar= \frac{1}{1.013}\ atm\).
The final pressure \(P_2 = 1.0\ millibar=1.0\times10^{- 3}\ bar\).
Converting to atm: \(P_2=1.0\times10^{-3}\times\frac{1}{1.013}\ atm\approx9.87\times10^{-4}\ atm\)

Step2: Apply the combined gas law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\), where \(P_1 = 0.90\ atm\), \(V_1=12.5\ L\), \(T_1 = 294\ K\), \(P_2\approx9.87\times 10^{-4}\ atm\), \(T_2 = 268\ K\)
We need to solve for \(V_2\), so \(V_2=\frac{P_1V_1T_2}{P_2T_1}\)
Substitute the values:
\(V_2=\frac{0.90\times12.5\times268}{9.87\times10^{-4}\times294}\)
First calculate the numerator: \(0.90\times12.5\times268 = 0.90\times3350=3015\)
Then calculate the denominator: \(9.87\times10^{-4}\times294\approx0.289\)
\(V_2=\frac{3015}{0.289}\approx1.0\times10^{4}\ L\)

Answer:

\(1.0\times 10^{4}\ L\) (the third option)