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Standard XII

Chemistry

Work Done in Isothermal Reversible Process

An inverted v...

Question

An inverted vessel (bell) lying at the bottom of a lake $50.6 m$ deep has $50 c c$ of air trapped in it. The bell is brought to the surface of lake. The volume of the trapped air will now be

200 c c

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250 c c

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300 c c

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350 c c

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Solution

The correct option is D $300 c c$
Let the $P_{o} = 10^{5} N / m^{2}$ ,
Assuming process to be isothermal , $P_{1} V_{1} = P_{2} V_{2}$
$(P_{0} + ρ g h) \times 50 = P_{0} V_{2}$
$(10^{5} + 1000 \times 10 \times 50.6) \times 50 = 10^{5} V_{2}$
$V_{2} = 300 c c$

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An inverted vessel (bell) lying at the bottom of a lake 50.6 m deep has 50 cc of air trapped in it. The bell is brought to the surface of lake. The volume of the trapped air will now be

The correct option is D 300 ccLet the Po=105 N/m2,Assuming process to be isothermal ,P1V1=P2V2 (P0+ρgh)×50=P0V2(105+1000×10×50.6)×50=105V2V2=300 cc

An inverted vessel (bell) lying at the bottom of a lake $50.6 m$ deep has $50 c c$ of air trapped in it. The bell is brought to the surface of lake. The volume of the trapped air will now be

The correct option is D $300 c c$
Let the $P_{o} = 10^{5} N / m^{2}$ ,
Assuming process to be isothermal , $P_{1} V_{1} = P_{2} V_{2}$
$(P_{0} + ρ g h) \times 50 = P_{0} V_{2}$
$(10^{5} + 1000 \times 10 \times 50.6) \times 50 = 10^{5} V_{2}$
$V_{2} = 300 c c$