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). 1.5×10² l 1.2 l 2.0×10⁴ l 1.0×10² l 1.0×10⁴ l

Explanation:

Step1: Convert pressure units

We know that \(1\ atm = 1.013\ bar\) and \(1\ bar= 1000\ mbar\).
The initial pressure \(P_1 = 0.90\ atm\). Convert it to millibar:
\(P_1=0.90\times1.013\times1000\ mbar = 911.7\ mbar\)

Step2: Use the combined gas law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\).
We are given \(V_1 = 12.5\ L\), \(T_1=294\ K\), \(P_1 = 911.7\ mbar\), \(T_2 = 268\ K\), \(P_2=1.0\ mbar\).
We need to solve for \(V_2\). Rearranging the combined - gas law formula gives \(V_2=\frac{P_1V_1T_2}{T_1P_2}\).
Substitute the values:
\(V_2=\frac{911.7\times12.5\times268}{294\times1.0}\)
First, calculate the numerator: \(911.7\times12.5\times268=(911.7\times12.5)\times268 = 11396.25\times268 = 3.054295\times10^{6}\)
Then, calculate the denominator: \(294\times1.0 = 294\)
\(V_2=\frac{3.054295\times 10^{6}}{294}\approx1.0\times10^{4}\ L\)

Answer:

\(1.0\times 10^{4}\ L\)