Question

The temperature of a given mass of air was changed from ${15}^{\circ }C$ to $-{15}^{\circ }C$. If the pressure remains unchanged and the initial volume was 100 mL, what should be the final volume?

• A. $V=89.58mL$
• B. $V=178.16mL$
• C. $V=192.79mL$
• D. None of these

Answer 1

The correct option is A $V=89.58mL$

Given
$T_1={15}^{\circ }C$ or $15+273.15K=288.15K$
$V_1=100mL$
$T_2=-{15}^{\circ }C=273.15-15=258.15K$
$V_2=?$
We know $\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$\therefore V_2=\frac{V_1}{T_1}\times T_2=\frac{100\times 258.15}{288.15}=89.58mL$