JEE Advanced Model Paper 1 with solutions for online practice

η =

Total‌work‌done

Total‌heat‌supplied

WAB = nR (TB−TC) = 600 R
WBC = nRTC ln (

P1

P2

) = 1200 R ln2
WCA = ∫ PdV , PV = nRT ,
PT = K
⇒

nRT

V

. T = K
⇒ nRT2 = KV
2nRT . dT = K . dv
∴ WCA = ∫

K

T

2nRTdT

K

=

300

∫

600

2nRdT = -1200 R
QAB=n CP ΔT = 2 ×

5

2

R × 300 = 1500 R
QBC = WBC = 1200 R ln2
QCA = dUCA + WCA = nCV dT + WCA = -900 R = Heat lost
Wnet = WAB + WBC + WCA = 1200 R ln2 - 600 R
Qabsorbed = QAB + QBC
∴ η =

600R(2ln2−1)

1500R+1200Rln2

= 1 -

21

15+12ln2

x = 7