Question

At constant temperature on increasing the pressure of a gas by $10\%,$ its volume will decrease by

• A. $9.09\%$
• B. $10\%$
• C. $5\%$
• D. $20\%$

Answer 1

Answer: (A)

Since, we know,
$p \propto \frac{1}{V}$
$\therefore \frac{p_{1}}{p_{2}}=\frac{V_{2}}{V_{1}}$
$\Rightarrow p_{2}=\frac{110}{100}p_{1} \, $ (given)
$\therefore \frac{V_{1}}{V_{2}}=\frac{p_{2}}{p_{1}}=\frac{110}{100}$
$\Rightarrow V_{2}=\frac{100}{110}V_{1}$
$\therefore $ Change in volume, $\Delta V=V_{2}-V_{1}$
$=\frac{100}{110}V_{1}-V_{1}=V_{1}\left(\frac{100 - 110}{110}\right)=\frac{- 10 V_{1}}{110}$
Negative sign means volume is decreased.
$\therefore \frac{\Delta V}{V_{1}}\times 100=\frac{10}{110}\times 100=9.09\%$