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 cm)3	...	48	60	...	100	...	200
p(in atmospheres)	2	...	\(\frac{3}{2}\)	1	...	\(\frac{1}{2}\)	...

Answer 1

 

v(in cm)3	45	48	60	90	100	180	200
p(in atmospheres)	2	\(\frac{15}{8}\)	\(\frac{3}{2}\)	1	0.9	\(\frac{1}{2}\)	\(\frac{9}{20}\)

Explanation: 

60 x \(\frac{3}{2}\) = x₁ x 48

60 x \(\frac{3}{2\times 48}\) = x₁

\(\frac{15}{8}\) = x₁

48 x \(\frac{15}{8}\) = x₂ x 2

48 x \(\frac{15}{8\times 2}\) = x²

45 = x²

60 x \(\frac{3}{2}\) = x₃ x 1

90 = x₃

90 x 1= x₄ x 100

\(\frac{90}{100}\) = x₄

0.9 = x₄

100 x 0.9 = x₅ x \(\frac{1}{2}\)

90 x 2 = x₅

180 = x₅

180 x \(\frac{1}{2}\) = x₆ x 200

\(\frac{90}{200}\) = x₆

\(\frac{9}{20}\) = x₆