An ideal gas is taken through a cyclic thermodynamical process through four steps $Q_{1}$ = 5960J, $Q_{2}$ = –5585J, $Q_{3}$ = –2980J and $Q_{4}$ = 3645J, respectively. The corresponding works involved are $W_{1} = 2200 J, W_{2} = - 825 J, W_{3} = - 1100 J$ and $W_{4}$ respectively. The value of W4 is:

1315J

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275J

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765J

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675J

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Solution

The correct option is C
765J

In a cyclic process $Δ U = 0$
1st law of thermodynamics
$Q = Δ U + W Q_{1} + Q_{2} + Q_{3} + Q_{4} = W_{1} + W_{2} + W_{3} + W_{4} 5960 - 5585 - 2980 + 3645 = 2200 - 825 - 1100 + W_{4} W_{4} = 1040 - 275 W_{4} = 765 J$

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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

and

Q_{4} = 3645 J

, respectively. The corresponding amounts of work done are

W_{1} = 2200 J, W_{2} = - 825 J

and

W_{3} = - 1100 J

and

W_{4} = r e s p e c t i v e l y

.
Find the value of

W_{4}

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

Q_{1} = 5960 J

Q_{2} = - 5585 J

Q_{3} = - 2980 J

and

Q_{4} = 3645 J

respectively. The corresponding works involved are

W_{1} = 2200 J

W_{2} = - 825 J

W_{3} = - 1100 J

and

W_{4}

respectively.
What is the efficiency of the cycle?

An ideal gas is taken through a cyclic thermodynamic process through four steps. The measures of heat involved in these steps are: $Q_{1} = 5960 J, Q_{2} = - 5585 J, Q_{3} = - 2980 J$ and $Q_{4} = 3645 J,$ respectively. The corresponding quantities of work involved are $W_{1} = 2200 J, = W_{2} = - 825 J$ and $W_{3} = 1100 J$ respectively.