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Difference between Internal Energy and Enthalpy

An ideal gas ...

Question

An ideal gas is subjected to a cyclic process involving four thermodynamic states; the amounts of heat (Q) and work (W) involved in each of these states are $Q_{1} = 6000 J, Q_{2} = 5500 J, Q_{3} = - 3000 J, Q_{4} = 3500 J W_{1} = 2500 J, W_{2} = - 1000 J, W_{3} = - 1200 J, W_{4} = x J$ . The ratio of the net work done by the gas to the total heat absorbed by the gas in ' $η$ '. The values of $x$ and $η$ respectively are:

A

500, 7.5 %

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B

1000, 10.5 %

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C

1000, 21 %

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D

1500, 15 %

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Solution

The correct option is C $1000, 10.5 %$
According to the data given in the question,
Net work done $= 2500 - 1000 - 1200 + 700$
$= 1000 J$
Heat absorbed $= 6000 + 3500$
$= 9500 J$
$η = \frac{1000}{9500} \times 100$
$= 10.5 %$

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Q. An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states are

Q_{1} = 6000 J,

Q_{2} = - 5500 J;

Q_{3} = - 3000 J;

Q_{4} = 3500 J

W_{1} = 2500 J;

W_{2} = - 1000 J;

W_{3} = - 1200 J;

W_{4} = x J .

The ratio of the net work done by the gas to the total heat absorbed by the gas is . The values of

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Q. An ideal gas after going through a series of four thermodynamic states in order reaches the initial state again (cyclic process) the amounts of heat (Q) and work (W) involved in the states are

Q_{1} = 6000 J

Q_{2} = - 5500 J

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W_{3} = - 1200 J

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Q. An ideal gas is taken through a cyclic thermodynamic process involving four steps. The amounts of heat involved in these steps are

Q_{1} = 5960 J, Q_{2} = - 5585 J, Q_{3} = - 2980 J

Q_{4} = 3645 J

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Q. An ideal gas is taken through a cyclic thermodynamic process involving four steps. The amounts of heat involved in these steps are

Q_{1} = 5960 J, Q_{2} = - 5585 J, Q_{3} = - 2980 J

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Q. An ideal gas is taken through a cyclic thermodynamic process through four steps

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