Question

For a $10\%$ increase in volume, what would be the $\%$ increase in temperature at a constant pressure?

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Apply Charle’s law; $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$V_1=V{V}_2=(V+\frac{10V}{100})=\frac{110V}{100}{T}_1=T{T}_2=x\frac{V}{T}=\frac{\left(\frac{110V}{100}\right)}{x};x=\frac{110V}{100}\times \frac{T}{V}\therefore T_2\frac{100T_1}{100}=1.1T_1\%increase=(1.1T_1-T_1)\times 100=10\%$