Question

a gas balloon has a volume of 9.9 liters when the temperature is 18.3 °c and the pressure is 755.3 mm of mercury. what will its volume be, in l, at 29.5 °c and 752.8 mm of mercury? express your answer with 1 decimal place.

Explanation:

Step1: Convert temperatures to Kelvin

$T_1 = 18.3 + 273.15=291.45\ K$, $T_2=29.5 + 273.15 = 302.65\ K$

Step2: Apply the combined - gas law $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

We know $P_1 = 755.3\ mmHg$, $V_1 = 9.9\ L$, $T_1 = 291.45\ K$, $P_2 = 752.8\ mmHg$, $T_2 = 302.65\ K$. Rearranging for $V_2$ gives $V_2=\frac{P_1V_1T_2}{T_1P_2}$

Step3: Substitute values and calculate

$V_2=\frac{755.3\times9.9\times302.65}{291.45\times752.8}$
$V_2=\frac{755.3\times9.9\times302.65}{291.45\times752.8}\approx10.3\ L$

Answer:

$10.3$