Question

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

• 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. $-1.0 \times 10^{7} \,Nm ^{-5}$

Answer 1

Answer: (C)

$PV ^{\gamma}=$ constant
$V ^{\gamma} \frac{ d P }{ dV }=\gamma PV ^{\gamma-1} \frac{ d V }{ dV }=0$
$\frac{ dP }{ dV }=\frac{-\gamma PV ^{\gamma-1}}{ V ^{\gamma}}=\frac{-\gamma P }{ V }$
$=-1.4 \times \frac{0.7 \times 10^{5}}{0.0049}$
$=-2 \times 10^{7}$