Answer:
Given \[{{V}_{1}}=120mL,\,\,\,\,\,\,\,\,\,{{P}_{1}}=1.2bar\]
\[{{V}_{2}}=180mL,\,\,\,\,\,\,\,\,\,{{P}_{2}}=?\]
Since, the
temperature is constant, Boyle's law will be applicable
\[{{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}\]
\[1.2\times 120={{P}_{2}}\times 180\]
\[{{P}_{2}}=\frac{1.2\times 120}{180}\]
\[=0.8\text{bar}\]
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