Half mole of an ideal gas (y=53) is taken through the cycle abcda as shown in figure. Take R = 253 JK−1mol−1
(a) Frnd the temperature of the gas in the states a, b, c, and d.
(b) Find the amount of heat supplied in the processes ab and bc.
(c) Find the amount of heat liberated in the processes cd and da.
n = 12mol.
R = 253 J \mol K
γ=53
(a) Temperature ata a= Ta
Pa Va = nRTa
Ta = PaVanR =120 K
Similary, temperature at b-240 K, At cit is 480 K and at d it is 240 K.
(b) For ab process
dQ = ncpdT
[Since ab is isometric]
= 12×Ryγ−1(Tb−Ta)
= 12×25×53×353−1×(240−120)
= 12×1259×32×(120)
= 1250 J
For bc, dQ = dU + dW
[dW = 0, Isochoric process]
dQ = dU = nCvdT
= nCv (tc-Tb)
= 12×(253)[(53−1)]×(240)
= 12×253×32×240=1500J
(c) Heat liberated in cd = - n CpdT
= - 12×γR(γ−1)×(TdTc)
= - 12×1259×32×(240−480)
= - 12×1256×240 = 2500 J.
Heat liberated in da
= - nCvdT
= -12×Rγ−1(Ta−Tb)
= -12×252×(120−240)
= 254×120−750 J