T P P 分≈T.0.57P=0257 2.27T (b) For the isochoric process a->b Q=nCr(T-T)=nCr (3T4-T,)=2nCpT For the isochoric process c->d Q2=nC(T-7)=nCV(T-2277)=-1.27nCT The efficiency of the cycle 4. One(1.00)mole of an ideal monatomic gas is taken around the cycle shown in Figure 5 (a)What is the work done by the gas in going from point ()to 2PoF---------. point(2)? and the work done by the gas in going from point(2)to (b)What is the change in the internal energy of the gas in going from PoF---i nt(1) to point3) along the path(1)→(2)→(3)? (c)What is the entropy change of the gas in going form point(1)to 2 point (3)? Fig 5 (d)What is the change in the internal energy and entropy of the gas in one complete cycle? Solution (a) The work done by the gas in going from point(1)to point(2)is the area under the curve(1)-) (2). That is W=Po(2Vo-V)=PoL The work done by the gas in going from point(2)to point (3)is zero b) Using the equation of state for an ideal gas: PV=nRT The temperature at point(I)is T- Pv, Pvo R R The temperature at point(3)is T, p=2P-2=4PVo R R R The change in the internal energy of the gas in going from point(1)to point (3)along the path(1) →(2)→(3)is AE=nC(T-T)=ir 4P._Po0)=PV=Poo R R 2a a a a c c d d c d c d P T T T P T T P P P T T 0.57 0.25 2.27 = ⇒ = ⋅ = ⋅ = (b) For the isochoric process a→b : Q nCV Tb Ta nCV Ta Ta 2nCVTa ( ) (3 ) 1 = − = − = For the isochoric process c→d : Q nCV Td Tc nCV Ta Ta 27nCVTa ( ) ( 2.27 ) 1. 2 = − = − = − The efficiency of the cycle is 0.365 2 1.27 1 1 1 2 = − = − = − v a v a nC T nC T Q Q η 4. One (1.00) mole of an ideal monatomic gas is taken around the cycle shown in Figure 5. (a) What is the work done by the gas in going from point (1) to point (2)? and the work done by the gas in going from point (2) to point (3)? (b)What is the change in the internal energy of the gas in going from point (1) to point (3) along the path (1) → (2) → (3)? (c) What is the entropy change of the gas in going form point (1) to point (3)? (d) What is the change in the internal energy and entropy of the gas in one complete cycle? Solution: (a) The work done by the gas in going from point (1) to point (2) is the area under the curve (1) → (2). That is 1 0 0 0 0 0 W = P (2V −V ) = PV The work done by the gas in going from point (2) to point (3) is zero. (b) Using the equation of state for an ideal gas: PV = nRT The temperature at point (1) is R PV R PV T 1 1 0 0 1 = = The temperature at point (3) is R PV R P V R PV T 3 3 0 0 0 0 3 2 2 4 = ⋅ = = The change in the internal energy of the gas in going from point (1) to point (3) along the path (1) → (2) → (3) is 0 0 0 0 0 0 0 0 3 1 2 3 2 ) 4 ( 2 ( ) PV PV i R PV R iR PV ∆E = nCV T − T = − = = 2V0 2P0 1 2 3 Fig.5 P0 P V V0