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Standard XII

Chemistry

Work done in Isothermal Irreversible Process

The three pro...

Question

The three processes in a thermodynamic cycle shown in the figure are : Process $1 \to 2$ is isothermal; Process $2 \to 3$ is isochoric (volume remains constant); Process $3 \to 1$ is adiabatic. The total work done by the ideal gas in this cycle is $10 J$ . The internal energy decreases by $20 J$ in the isochoric process. The work done by the gas in the adiabatic process is $- 20 J$ . The heat added to the system in the isothermal process is

0 J

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10 J

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20 J

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30 J

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Solution

The correct option is D $30 J$
Total work done in isothermal process is =10+20=30 J
Since in isothermal there is no change is internal energy, So all heat is converted to work. So the heat added to system is 30 J

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The correct option is D 30JTotal work done in isothermal process is =10+20=30 JSince in isothermal there is no change is internal energy, So all heat is converted to work. So the heat added to system is 30 J

The correct option is D $30 J$
Total work done in isothermal process is =10+20=30 J
Since in isothermal there is no change is internal energy, So all heat is converted to work. So the heat added to system is 30 J