Q.
3 moles of an ideal monoatomic gas performs ABCDA cyclic process as shown in figure below. The gas temperatures are TA=400K, TB=800K,TC=2400K and TD=1200K. The work done by the gas is (approximately) (R=8.314J/molK)
Processes A to B and C to D are parts of straight line graphs of form y=mx.
Also, p=VμRT(μ=3) p∝T
So, volume remains constant for the graphs AB and CD
So, no work is done during processes for A to B and C to D. WAB=WCD=0
and WBC=p2(VC−VB) =μR(TC−TB) =3R(2400−800) =3R×1600 =4800R WDA=p1(VA−VD) =μR(TA−TD) =3R(400−1200) =−2400R
Work done in complete cycle W=WAB+WBC+WCD+WDA =0+4800R+0+(−2400)R =2400R =19.944J=20kJ