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JEE Main 2013 · P-V Diagrams & Cyclic Processes

P-V Diagrams & Cyclic Processes question from JEE Main 2013

Q80JEE Main 2013Difficult

P-V Diagrams & Cyclic Processes

Figure

The above p-vp\text{-}v diagram represents the thermodynamic cycle of an engine, operating with an ideal monatomic gas. The amount of heat, extracted from the source in a single cycle is

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Answer: (b)

  1. nRT=PVnRT = PV: D=p0v0D = p_0v_0, A=2p0v0A = 2p_0v_0, B=4p0v0B = 4p_0v_0, C=2p0v0C = 2p_0v_0. Heat is taken in only where TT rises: D→AD \to A and A→BA \to B.
  2. D→AD \to A (isochoric): Q=32nRΔT=32(2p0v0−p0v0)=32p0v0Q = \frac{3}{2}nR\Delta T = \frac{3}{2}(2p_0v_0 - p_0v_0) = \frac{3}{2}p_0v_0
  3. A→BA \to B (isobaric): Q=52nRΔT=52(4p0v0−2p0v0)=5p0v0Q = \frac{5}{2}nR\Delta T = \frac{5}{2}(4p_0v_0 - 2p_0v_0) = 5p_0v_0
  4. Heat from source =32p0v0+5p0v0=132p0v0= \frac{3}{2}p_0v_0 + 5p_0v_0 = \frac{13}{2}p_0v_0. So the answer is (b).
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