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, $\Delta V=V_2-V_1$
$=\frac{100}{110}{V}_1-V_1=V_1\left(\frac{100-110}{110}\right)=\frac{-10V_1}{110}$
Negative sign means volume is decreased.
$\therefore \frac{\Delta V}{V_1}\times 100=\frac{10}{110}\times 100=9.09\%$