Process undergone by
1 mole gas is as shown :
For an ideal monoatomic gas, ratio of specific heat
γ= For adiabatic process
A→B pAVAγ=pBVBγ Substituting values of
pA,pB and
VB from graph,
⇒32×105×VAγ=1×105×(1)γ ⇒VAγ=()γ Here,
γ= ⇒(VA)= or
VA=()==m3 Similarly for adiabatic compression
C→D,
pCVCγ=pDVDγ We substitute values from graph,
⇒(1×105)×(8)=(32×105)(VD) ⇒VD5∕3=1 ⇒VD=13∕5=1m3 Now, Work done in process
A→B→C→D→A =WAB+WBC+WCD+WDA Here we substitute work done for adiabatic process,
Wadiabatic= and work done of isobaric process,
Wisobaric=p(Vf−Vi) so, work done in given cycle
ABCDA =WAB+WBC+WCD+WDA =+pB(VC−VB)+()+pD(VA−VD) Substituting values of pressures and volumes, we get
WABCDA=(| 32×105×−1×105×1 |
| −1 |
)+1×105(8−1)+(| 1×105×8−32×105×1 |
| −1 |
) +32×105(−1) =+7×105++4×105×x−7=−52.5×105J