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Question

An ideal gas (γ = 1.67) is taken through the process abc shown in the figure. The temperature at point a is 300 K. Calculate (a) the temperatures at b and c (b) the work done in the process (c) the amount of heat supplied in the path ab and in the path bc and (d) the change in the internal energy of the gas in the process.

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Solution

(a) For line ab, volume is constant.
So, from the ideal gas equation,
P1T1=P2T2100300=200T2



T2=200×300100=600 K

For line bc, pressure is constant.
So, V1T1=V2T2100600=150T2T2 600×150100=900 K

(b)
As process ab is isochoric, Wab=0.
During process bc,
P = 200 kPa

The volume is changing from 100 to 150 cm3 .
Therefore, work done = 50 × 10−6 × 200 × 103 J
= 10 J

(c) For ab (isochoric process), work done = 0.
From the first law,
dQ = dU = nCvdT
Heat supplied = nCvdT
Now,
Qab=PVRT×Rγ-1×dT=200×103×100×10-6×300600×0.67=14.925 (γ=1.67)For bc (isobaric process):Heat supplied in bc=nCpdT Cp=γRγ-1=PVRT×γRγ-1×dT=200×103×150×10-6600×0.67×300=10×1.670.67=16.70.67=24.925

(d) dQ = dU + dW
Now,
dU = dQ − dW
= Heat supplied − Work done
= (24.925 + 14.925) − 10
= 39.850 − 10 = 29.850 J.

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