Question Number 35197 by Tinkutara last updated on 16/May/18
Answered by ajfour last updated on 17/May/18
$$ \\ $$$$\Rightarrow\:{P}\:=\:\frac{\mathrm{3}×\mathrm{320}}{\mathrm{290}}\:{atm}\:=\frac{\mathrm{96}}{\mathrm{29}}{atm} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{96}}{\mathrm{30}−\mathrm{1}}=\frac{\mathrm{96}}{\mathrm{30}}\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{30}}\right)\: \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{3}.\mathrm{2}+\frac{\mathrm{3}.\mathrm{2}}{\mathrm{30}}\:=\:\mathrm{3}.\mathrm{31}{atm} \\ $$$$\frac{{n}−\bigtriangleup{n}}{\mathrm{3}}=\frac{{n}}{\mathrm{3}.\mathrm{31}} \\ $$$$\Rightarrow\:\:\mathrm{1}−\frac{\bigtriangleup{n}}{{n}}=\frac{\mathrm{3}}{\mathrm{3}.\mathrm{31}} \\ $$$${or}\:\:\:\frac{\bigtriangleup{n}}{{n}}=\frac{\mathrm{0}.\mathrm{31}}{\mathrm{3}.\mathrm{31}}\:=\frac{\bigtriangleup{V}\mid_{\mathrm{3}.\mathrm{31}{atm}} }{{V}\mid_{\mathrm{3}.\mathrm{31}{atm}} =\mathrm{10}\:{l}} \\ $$$$\Rightarrow\:\left(\bigtriangleup{V}\right)_{\mathrm{1}{atm}} =\left(\bigtriangleup{V}\right)_{\mathrm{3}.\mathrm{31}{atm}} ×\mathrm{3}.\mathrm{31} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{10}×\frac{\mathrm{0}.\mathrm{31}}{\mathrm{3}.\mathrm{31}}×\mathrm{3}.\mathrm{31} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{3}.\mathrm{1}\:{litres}. \\ $$
Commented by Tinkutara last updated on 16/May/18
Answer of (b) part is not correct Sir.
Commented by ajfour last updated on 17/May/18
what's the correct answer?
Commented by Tinkutara last updated on 17/May/18
Your answer is correct Sir. Thanks a lot!