Question

It is known that for a given mass of gas, the volume v varies inversely as the pressure p. Fill in the missing entries in the following table:

v (in cm3)	...	48	60	...	100	...	200
p (in atmospheres)	2	...	3/2	1	...	1/2	...

Answer 1

$$\begin{gathered} \mathrm{Since}\mathrm{the}\mathrm{volume}\mathrm{and}\mathrm{pressure}\mathrm{for}\mathrm{the}\mathrm{given}\mathrm{mass}\mathrm{vary}\mathrm{inversely},\mathrm{we}\mathrm{have}: \\ vp=k \\ \mathrm{For}v=60\mathrm{and}p=\frac{3}{2},\mathrm{we}\mathrm{have}: \\ k=60\times \frac{3}{2} \\ =90 \\ \mathrm{For}p=2\mathrm{and}k=90,\mathrm{we}\mathrm{have}: \\ 2v=90 \\ \Rightarrow v=\frac{90}{2} \\ =45 \\ \mathrm{For}v=48\mathrm{and}k=90,\mathrm{we}\mathrm{have}: \\ 48p=90 \\ \Rightarrow p=\frac{90}{48} \\ =\frac{15}{8} \\ \mathrm{For}p=1\mathrm{and}k=90,\mathrm{we}\mathrm{have}: \\ 1v=90 \\ \Rightarrow v=\frac{90}{1} \\ =90 \\ \mathrm{For}v=100\mathrm{and}k=90,\mathrm{we}\mathrm{have}: \\ 100p=90 \\ \Rightarrow v=\frac{90}{100} \\ =\frac{9}{10} \\ \mathrm{For}p=\frac{1}{2}\mathrm{and}k=90,\mathrm{we}\mathrm{have}: \\ \frac{1}{2}v=90 \\ \Rightarrow v=90\times 2 \\ =180 \\ \mathrm{For}v=200\mathrm{and}k=90,\mathrm{we}\mathrm{have}: \\ 200p=90 \\ \Rightarrow p=\frac{90}{200} \\ =\frac{9}{20} \end{gathered}$$