Answer:
Information shadow:
At \[23.4~{{\,}^{{}^\circ
}}C\]temperature, the volume of the balloon is\[{{26.1}^{{}^\circ }}C\]. The temperature
of Indian ocean is, the volume at this temperature is to be calculated.
Problem
solving strategy:
At constant
pressure, Charles' law can be applied
\[\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}\]
Working
it out:
\[{{T}_{1}}=23.4+273=296.4K\]
\[{{V}_{1}}=2L\]
\[{{T}_{2}}=26.1+273=299.1K\]
\[{{V}_{2}}=?\]
\[\frac{{{V}_{1}}}{{{T}_{1}}}=\frac{{{V}_{2}}}{{{T}_{2}}}\]
\[\frac{2}{296.4}=\frac{{{V}_{2}}}{299.1}\]
\[{{V}_{2}}=2.018L\]
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