Question

1. a gas sample at 40.0°c occupies a volume of 2.32 l. if the temperature is raised to 75.0°c, what will the volume be, assuming the pressure remains constant?
2. a gas at 89°c occupies a volume of 0.67 l. at what celsius temperature will the volume increase to 1.12 l?
3. the celsius temperature of a 3.00 - l sample of gas is lowered from 80.0°c to 30.0°c. what will be the resulting volume of this gas?
4. what is the volume of the air in a balloon that occupies 0.620 l at 25°c if the temperature is lowered to 0.00°c?
5. calculate the decrease in temperature when 2.00 l at 20.0 °c is compressed to 1.00 l.
6. 600.0 ml of air is at 20.0 °c. what is the volume at 60.0 °c?
7. a gas occupies 900.0 ml at a temperature of 27.0 °c. what is the volume at 132.0 °c?
8. what change in volume results if 60.0 ml of gas is cooled from 33.0 °c to 5.00 °c?
9. given 300.0 ml of a gas at 17.0 °c. what is its volume at 10.0 °c?

Explanation:

1.

Step1: Convertir temperaturas a Kelvin

$T_1 = 40.0 + 273.15=313.15\ K$, $T_2 = 75.0 + 273.15 = 348.15\ K$, $V_1 = 2.32\ L$
Según la ley de Charles ($\frac{V_1}{T_1}=\frac{V_2}{T_2}$), despejamos $V_2$.

Step2: Calcular $V_2$

$V_2=\frac{V_1\times T_2}{T_1}=\frac{2.32\times348.15}{313.15}\approx2.57\ L$

Step1: Convertir temperatura inicial a Kelvin

$T_1 = 89+ 273.15 = 362.15\ K$, $V_1 = 0.67\ L$, $V_2 = 1.12\ L$
Usamos la ley de Charles $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ y despejamos $T_2$.

Step2: Calcular $T_2$ en Kelvin

$T_2=\frac{V_2\times T_1}{V_1}=\frac{1.12\times362.15}{0.67}\approx604.98\ K$

Step3: Convertir $T_2$ a Celsius

$t_2=T_2 - 273.15=604.98 - 273.15 = 331.83^{\circ}C$

Step1: Convertir temperaturas a Kelvin

$T_1 = 80.0+273.15 = 353.15\ K$, $T_2 = 30.0 + 273.15=303.15\ K$, $V_1 = 3.00\ L$
Aplicamos la ley de Charles $\frac{V_1}{T_1}=\frac{V_2}{T_2}$ y despejamos $V_2$.

Step2: Calcular $V_2$

$V_2=\frac{V_1\times T_2}{T_1}=\frac{3.00\times303.15}{353.15}\approx2.57\ L$

Answer:

$2.57\ L$

2.