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← JEE Advanced 2026 · Paper 1
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JEE Advanced 2026 · Paper 1, Physics question 7

A quasi-static cycle of a monoatomic ideal gas contains an isothermal process (ab\boldsymbol{ab}), followed by an isochoric process (bc\boldsymbol{bc}) and an adiabatic process (ca\boldsymbol{ca}) as shown in the figure. The volumes of the gas are V1V_1 and V2V_2 at a\boldsymbol a and b\boldsymbol b, respectively. If the cycle has heat input QinQ_{\text{in}} and output QoutQ_{\text{out}}, then the efficiency of the cycle is defined as η=Qin−QoutQin\eta=\dfrac{Q_{\text{in}}-Q_{\text{out}}}{Q_{\text{in}}}. The correct statement(s) is/are:
[Given: ln⁡2≈0.7\ln2\approx0.7]

P-V diagram: from a (at V₁) an isothermal curve runs to b (at V₂); bc drops vertically at V₂ to c; the steeper adiabat ca returns to a. Arrows show the cycle a → b → c → a.
Choose one or more options
Answer and worked solution

Official answer: A, B, C

Monoatomic gas, γ=53\gamma=\frac53. Let ThT_h be the isotherm temperature and r=V2/V1r=V_2/V_1.

Temperatures. The adiabat caca joins V2V_2 to V1V_1: TcV22/3=ThV12/3T_cV_2^{2/3}=T_hV_1^{2/3}, so ThTc=r2/3\dfrac{T_h}{T_c}=r^{2/3}. For r=8r=8 this is 44. (C) true.

Heats (per mole).

  • Absorbed on abab: Qab=RThln⁡rQ_{ab}=RT_h\ln r.
  • Released on bcbc: Qbc=32R(Th−Tc)=32RTh(1−r−2/3)Q_{bc}=\frac32R(T_h-T_c)=\frac32RT_h\left(1-r^{-2/3}\right).

(A) For r=8r=8: Qab=3(0.7)RTh=2.1RThQ_{ab}=3(0.7)RT_h=2.1RT_h and Qbc=98RTh≈1.13RThQ_{bc}=\frac98RT_h\approx1.13RT_h. Released < absorbed. True.

(B) η=1−32(1−r−2/3)ln⁡r\eta=1-\dfrac{\frac32\left(1-r^{-2/3}\right)}{\ln r} depends only on rr. True.

(D) On the isotherm PaV1=PbV2P_aV_1=P_bV_2, so Pa/Pb=8P_a/P_b=8, not 4. False.