Question

The pressure of an ideal gas undergoing isothermal change is increased by 10%. The volume of the gas must decrease by about:

• A. $0.1\%$
• B. $9\%$
• C. $10\%$
• D. $0.9\%$

Answer 1

Answer: (B)

$9\%$

Change in pressure, $\Delta P=\frac{10}{100}\times p=0.1p$

Therefore, p + $\Delta p=1.1p$

If the process is carried out isothermally, then T = constant.

$P_1{V}_1=P_2{V}_2$

$p\times V_1=1.1p\times V_2$

Now. $V_2-V_1=\frac{V_1}{1.1}-V_1$

$=(\frac{1}{1.1}-1)V_1$

$=-\frac{0.1}{1.1}{V}_1=-0.09V_1$

There is a decrease in volume.

Hence, change in volume $=\frac{|V_2-V_1|}{V_1}\times 100$

$=9$%

Hence, option B is correct.