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(B) For an ideal gas,the equation of state is $PV = nRT$.For a constant temperature (isothermal) process,$T$ is constant. Since $n$ and $R$ are also c

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(B) For an ideal gas,the equation of state is $PV = nRT$.For a constant temperature (isothermal) process,$T$ is constant. Since $n$ and $R$ are also constants,the product $nRT$ is a constant,let's call it $k$.So,the equation becomes $PV = k$,which can be rewritten as $P = k \left( \frac{1}{V} \right)$.This equation is of the form $y = mx$,where $y = P$,$x = \frac{1}{V}$,and the slope $m = nRT$ is a positive constant.Therefore,the graph of $P$ versus $\frac{1}{V}$ is a straight line passing through the origin.

For an ideal gas undergoing a constant temperature process (isothermal process),what is the nature of the graph between pressure $(P)$ and $\frac{1}{V}$ (where $V$ is volume)?

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