Question

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

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

Answer 1

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

For reversible adiabat,
$P{v}^{\gamma }=constant\Rightarrow vdP+P\gamma dv=0\Rightarrow \frac{dP}{dv}=-\frac{\gamma P}{v}$
For $P=0.7\times {10}^{5}N{m}^{-2},v=0.0049m^3,\gamma =1.4$
required slope= $-\frac{1.4\times 0.7\times {10}^{5}N{m}^{-2}}{0.0049m^3}=-2\times {10}^{7}N{m}^{-5}$