Answer:
Information shadow:
The pressure that
the balloon can withhold is 0.2 bar. The volume of \[{{H}_{2}}\] gas at 1 bar
pressure is 2.27 L. Temperature is kept constant.
Problem
solving strategy:
At constant
temperature, Boyle's law will be applied i.e., \[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]
Working
it out:
\[{{P}_{1}}=1\]bar \[{{V}_{1}}=2.27L\]
\[{{P}_{2}}=0.2\]bar \[{{V}_{2}}=?\]
\[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]
\[1\times
2.27=0.2\times {{V}_{2}}\]
\[{{V}_{2}}=11.35L\]
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