Answer:
\[{{P}_{1}}{{V}_{1}}+{{P}_{2}}{{V}_{2}}={{P}_{3}}{{V}_{3}}\] .(i)
\[{{P}_{1}}=0.8bar,\] \[{{V}_{1}}=0.5L\]
\[{{P}_{2}}=0.7bar,\] \[{{V}_{2}}=2L\]
\[{{P}_{3}}=?\] \[{{V}_{3}}=1L\]
Putting these values
in Eqn. (i)
\[0.8\times 0.5+0.7\times 2={{P}_{3}}\times
1\]
\[{{P}_{3}}=1.8\,\,bar\]
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