Question

Which of the following graphs correctly represents the variation of $\beta=-\left(\frac{dV}{dP}\right) \frac{1}{V}$ with $P$ for an ideal gas at constant temperature?

• A.
• B.
• C.
• D.

Answer 1

Answer: (A)

For an ideal gas, the differential from of gas law is $VdP + PdV = nRdT$
At constant temperature $dT = 0$
$\Rightarrow $ $VdP + PdV = 0$
$\frac{-dV}{dP}=\left(\frac{V}{P}\right) or-\left(\frac{dV}{dP}\right) \frac{1}{V}=\frac{1}{P} \Rightarrow \beta=\frac{1}{P}$
$\beta P = constant$
Hence the graph could be a rectangular hyperbola as shown in (a).