Question

A reversible adiabatic path on a $P-V$ diagram for an ideal gas passes through state $A$ where $P=0.7\times {10}^5{\text{N/m}}^{–2}$ and $V=0.0049{\text{m}}^3$. The ratio of specific heat of the gas is $1.4$. The slope of path at $A$ is

• A. $2.0\times {10}^7{\text{Nm}}^{-5}$
• B. $1.0\times {10}^7{\text{Nm}}^{-5}$
• C. $-2.0\times {10}^7{\text{Nm}}^{-5}$
• D. $-1.0\times {10}^7{\text{Nm}}^{-5}$

Answer 1

The correct option is C $-2.0\times {10}^7{\text{Nm}}^{-5}$

For reversible adiabatic,
$P{V}^{\gamma }=$ constant
$V^{\gamma }\frac{dP}{dV}=\gamma P{V}^{\gamma -1}\frac{dV}{dV}=0$
$\frac{dP}{dV}=\frac{-\gamma P{V}^{\gamma -1}}{V^{\gamma }}=\frac{-\gamma P}{V}$
For $P=0.7\times {10}^5{\text{Nm}}^{-2}$
$V=0.049{\text{m}}^3,\gamma =1.4$
Slope $=-1.4\times \frac{0.7\times {10}^5}{0.0049}=-2\times {10}^7$