Question

charles law problems: $\frac{v_1}{t_1}=\frac{v_2}{t_2}$ $v_1t_2 = v_2t_1$ 1 atm = 760.0 mm hg = 101.3 kpa 5. calculate the decrease in temperature when 6.00 l at 20.0 $^{circ}$c is compressed to 4.00 l.

Explanation:

Step1: Convert temperatura inicial a Kelvin

$T_1=(20.0 + 273)\text{ K}=293\text{ K}$

Step2: Aplicar ley de Charles ($V_1T_2 = V_2T_1$)

$T_2=\frac{V_2T_1}{V_1}$
Sustituir $V_1 = 6.00\text{ L}$, $V_2 = 4.00\text{ L}$ y $T_1 = 293\text{ K}$
$T_2=\frac{4.00\text{ L}\times293\text{ K}}{6.00\text{ L}}\approx195.33\text{ K}$

Step3: Calcular la disminución en temperatura

$\Delta T=T_1 - T_2$
$\Delta T=293\text{ K}- 195.33\text{ K}=97.67\text{ K}$

Answer:

$97.67\text{ K}$