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

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