Question

Calculate the temperature at which the volume of a given mass of gas gets reduced to 3/5th of original volume at 10${}^{o}C$ without any change in pressure.

• A. -171${}^{o}C$
• B. 513${}^{o}C$
• C. -198${}^{o}C$
• D. -103.2${}^{o}C$

Answer 1

The correct option is D -103.2${}^{o}C$

Given that original volume $V_1$ = V and final volume $V_2$ = $\frac{3}{5}V$
Also $T_1$ = 273 + 10 =283 $K$
Applying Charle's law; $\frac{V_1}{T_1}=\frac{V_2}{T_2}$
$T_2=\frac{3\times 283}{5}$ = 169.8 $K$
Thus $T_2$ = 169.8 - 273 = -${103.2}^{\circ }C$