Charles's Law Calculations 1. A container holds 50 mL of nitrogen at 25^circ C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 10^circ C i 2. A sample of oxygen occupies a volume of 160cm^3 at 91^circ C What will be the volume of oxygen when the temperature drops to 0^circ C i 3. A sample of hydrogen gas has an initial temperature of 50^circ C When the temperature is lowered to -5^circ C the volume of hydrogen becomes 212cm^3 What was the initial volume of the hydrogen gas? 4. 568cm^3 of chlorine at 25^circ C will occupy what volume at -25^circ C while the pressure remains constant? (V_(1))/(T_(1))=(V_(2))/(T_(2))

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Charles's Law Calculations 1. A container holds 50 mL of nitrogen at 25^circ C and a pressure of 736 mm Hg. What will be its volume if the temperature increases by 10^circ C i 2. A sample of oxygen occupies a volume of 160cm^3 at 91^circ C What will be the volume of oxygen when the temperature drops to 0^circ C i 3. A sample of hydrogen gas has an initial temperature of 50^circ C When the temperature is lowered to -5^circ C the volume of hydrogen becomes 212cm^3 What was the initial volume of the hydrogen gas? 4. 568cm^3 of chlorine at 25^circ C will occupy what volume at -25^circ C while the pressure remains constant? (V_(1))/(T_(1))=(V_(2))/(T_(2))

Answer 1

1. To solve this problem, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature, provided the pressure and the amount of gas remain constant. The formula for Charles's Law is: $\frac{V_1}{T_1} = \frac{V_2}{T_2}$ where $V_1$ is the initial volume, $T_1$ is the initial temperature, $V_2$ is the final volume, and $T_2$ is the final temperature. Given: $V_1 = 50 \text{ mL}$ $T_1 = 25^{\circ}C$ $T_2 = 35^{\circ}C$ We need to find $V_2$. First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15: $T_1 = 25^{\circ}C + 273.15 = 298.15 \text{ K}$ $T_2 = 35^{\circ}C + 273.15 = 308.15 \text{ K}$ Now, we can plug the values into the formula and solve for $V_2$: $\frac{50 \text{ mL}}{298.15 \text{ K}} = \frac{V_2}{308.15 \text{ K}}$ $V_2 = \frac{50 \text{ mL} \times 308.15 \text{ K}}{298.15 \text{ K}}$ $V_2 \approx 51.8 \text{ mL}$ Therefore, the volume of nitrogen will be approximately 51.8 mL when the temperature increases by $10^{\circ}C$. 2. Given: $V_1 = 160 \text{ cm}^3$ $T_1 = 91^{\circ}C$ $T_2 = 0^{\circ}C$ We need to find $V_2$. First, we need to convert the temperatures from Celsius to Kelvin: $T_1 = 91^{\circ}C + 273.15 = 364.15 \text{ K}$ $T_2 = 0^{\circ}C + 273.15 = 273.15 \text{ K}$ Now, we can plug the values into the formula and solve for $V_2$: $\frac{160 \text{ cm}^3}{364.15 \text{ K}} = \frac{V_2}{273.15 \text{ K}}$ $V_2 = \frac{160 \text{ cm}^3 \times 273.15 \text{ K}}{364.15 \text{ K}}$ $V_2 \approx 117.6 \text{ cm}^3$ Therefore, the volume of oxygen will be approximately 117.6 cm$^3$ when the temperature drops to $0^{\circ}C$. 3. Given: $V_2 = 212 \text{ cm}^3$ $T_1 = 50^{\circ}C$ $T_2 = -5^{\circ}C$ We need to find $V_1$. First, we need to convert the temperatures from Celsius to Kelvin: $T_1 = 50^{\circ}C + 273.15 = 323.15 \text{ K}$ $T_2 = -5^{\circ}C + 273.15 = 268.15 \text{ K}$ Now, we can plug the values into the formula and solve for $V_1$: $\frac{V_1}{323.15 \text{ K}} = \frac{212 \text{ cm}^3}{268.15 \text{ K}}$ $V_1 = \frac{212 \text{ cm}^3 \times 323.15 \text{ K}}{268.15 \text{ K}}$ $V_1 \approx 256.8 \text{ cm}^3$ Therefore, the initial volume of the hydrogen gas was approximately 256.8 cm$^3$. 4. Given: $V_1 = 568 \text{ cm}^3$ $T_1 = 25^{\circ}C$ $T_2 = -25^{\circ}C$ We need to find $V_2$. First, we need to convert the temperatures from Celsius to Kelvin: $T_1 = 25^{\circ}C + 273.15 = 298.15 \text{ K}$ $T_2 = -25^{\circ}C + 273.15 = 248.15 \text{ K}$ Now, we can

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