InterviewSolution
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InterviewSolution
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The above p-v diagram represents the thermodynamic cycle of an engine, operating with an ideal monoatomic gas. The amount of heat, extracted from the source in a single cycle isA. `P_(0)V_(0)`B. `(13/2)P_(0)V_(0)`C. `(11/2)P_(0)V_(0)`D. `4P_(0)V_(0)` |
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Answer» Correct Answer - B In the (figure)shown, suppose temp. of monoatomic gas at A is `T_(0)` In going A to B , volume is constant. As pressure increases from `P_(0)` to `2P_(0)` temp. increases from `T_(0)` to `2T_(0)` `:.` Heat extracted from source `= 1xx C_(v).dT` `Q_(1)=3/2R(2T_(0)-T_(0))= 3/2RT_(0)= 3/2P_(0)V_(0)` In going from B to C, pressure is constant. As volume increases from `V_(0) to 2 V_(0)`, temp. increases from `2T_(0) to 4T_(0) ( :. T prop V)` `:.` Heat extracted from source `=1xxc_(p).dT` `Q_(2)=1xx5/2R(4T_(0)-2T_(0))= 5RT_(0)= 5P_(0)V_(0)` In going from C to D, volume is constant. Pressure decreases from `2p_(0) to P_(0)`. Temp. decreases from `4T_(0) to 2T_(0)`. No heat is extracted from the source. In going to D to A, pressure is constant Volume decreases from `2v_(0) to V_(0)`. No heat is extracted from the source. Hence in single cycle, heat extracted from the source `=Q_(1)+Q_(2)= 3/2P_(0)V_(0)+5P_(0)V_(0)= (13)/2p_(0)v_(0)` |
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