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A thermodynamic system undergoes cyclic process ABCDA as shown in figure. The work done by the system in the cycle is |
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Answer» `P_(0)V_(0)` As is CLEAR from figure, `W_(AEDA) = + "area of " DELTA AED = + (1)/(2)P_(0)V_(0)` `W_(BCEB) = -"Area of "Delta BCE = -(1)/(2)P_(0)V_(0)` The net work done by the system is `W_("net") = W_(AEDA) + W_(BCEB) = + (1)/(2)P_(0)V_(0) - (1)/(2)P_(0)V_(0)` = zero
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