Question

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

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

Answer 1

Answer: (C)

$-2.0\times {10}^7{Nm}^{-5}$

We have,

$P=0.7\times {10}^{5}N/m^{-2}$

$V=0.0049m^3$

$\gamma =1.4$

For $P-V$ curve adiabatic equation is

$P{V}^{\gamma }=constant$ $............\left(1\right)$

On differentiating w.r.t $V$, we get

$\frac{dP}{dV}{V}^{\gamma }+\gamma P{V}^{\gamma -1}=0$

$\frac{dP}{dV}=-\frac{\gamma P{V}^{\gamma -1}}{V^{\gamma }}$

$\frac{dP}{dV}=-\frac{1.4\times (0.7\times {10}^5)\times (0.0049{)}^{1.4-1}}{(0.0049{)}^{1.4}}$

$\frac{dP}{dV}=-\frac{(0.98\times {10}^5)\times (0.0049{)}^{0.4}}{(0.0049{)}^{1.4}}$

$\frac{dP}{dV}=-\frac{0.98\times {10}^5}{(0.0049{)}^1}$

$\frac{dP}{dV}=-200\times {10}^5$

$\frac{dP}{dV}=-2\times {10}^{7}N{m}^{-5}$

Hence, this is the answer.