Areversible adiabatic path on a $P-V$ diagram for an ideal gas passes through stateAwhere $P=0$.$7\times 10^5 \,\,N/ m^{-2}$ and $v = 0.0049 \,\,m^3$. The ratio of specific heat of the gas is $1.4$. The slope of path at $A$ is :
Diffcult
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For reversible adiabat,
$P v^{\gamma}=$ constant $\Rightarrow v d P+P \gamma d v=0 \Rightarrow \frac{d P}{d v}=-\frac{\gamma P}{v}$
For $P=0.7 \times 10^{5} N m^{-2}, v=0.0049 m^{3}, \gamma=1.4$
required slope $=-\frac{1.4 \times 0.7 \times 10^{5} N m^{-2}}{0.0049 m^{3}}=-2 \times 10^{7} N m^{-5}$
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