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The work done in taking an ideal gas through one cycle of operation as shown in the indicator diagram below
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1605 Upvotes
Verified Answer
The correct answer is:
$10 \mathrm{~J}$
As work done in a cycle is equal to the area enclosed by the cycle.
$\therefore \quad$ Work done, $W=\Delta p \times \Delta V$
$$
=(4-2) \times(6-1)=10 \mathrm{~J}
$$
$\therefore \quad$ Work done, $W=\Delta p \times \Delta V$
$$
=(4-2) \times(6-1)=10 \mathrm{~J}
$$
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