AP EAMCET Engineering 17 Sep 2020 Shift 1 Paper

For an isothermal change,
pV=K
Differentiating both side, we get
∴ Slope of an isothermal curve
(‌

dp

dV

)‌iso ‌=−‌

p

V

For an adiabatic change,
pVγ=K′
Differentiating both side, we get
p⋅γVγ−1⋅dV+Vγ⋅dp=0
⇒‌‌γpdV+Vdp=0‌‌ or ‌‌Vdp=−γ⋅pdV
∴ Slope of an adiabatic curve,
(‌

dp

dV

)‌adia ‌=−‌

γp

V

∴‌‌‌

( dp dV )i50

(‌ dp dV )adia

=‌

−p/V

−γp/V

=‌

1

γ