Amonton's Law Calculations 1. Determine the pressure change when a constant volume of a gas at 101.3 kPa is heated from 25^circ C to 35^circ C 2. A gas has a pressure of 794 mm Hg at 43^circ C What is the temperature at standard pressure? 3. If a gas in a closed container is pressurized from 12.0 atm to 25.0 atm and its origir temperature was 32.0^circ C what would the final temperature of the gas be? 4. A sample of gas at 3000 mm Hg inside a steel tank is cooled from 500^circ C to 0^circ What is the final pressure of the gas in the steel tank? (P_(1))/(T_(1))=(P_(2))/(T_(2))

Question

Problemas

Amonton's Law Calculations 1. Determine the pressure change when a constant volume of a gas at 101.3 kPa is heated from 25^circ C to 35^circ C 2. A gas has a pressure of 794 mm Hg at 43^circ C What is the temperature at standard pressure? 3. If a gas in a closed container is pressurized from 12.0 atm to 25.0 atm and its origir temperature was 32.0^circ C what would the final temperature of the gas be? 4. A sample of gas at 3000 mm Hg inside a steel tank is cooled from 500^circ C to 0^circ What is the final pressure of the gas in the steel tank? (P_(1))/(T_(1))=(P_(2))/(T_(2))

Answer 1

1. To determine the pressure change when a constant volume of a gas at 101.3 kPa is heated from $25^{\circ }C$ to $35^{\circ }C$, we can use the formula for Amonton's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ where $P_1$ is the initial pressure, $T_1$ is the initial temperature in Kelvin, $P_2$ is the final pressure, and $T_2$ is the final temperature in Kelvin. First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15: $T_1 = 25^{\circ }C + 273.15 = 298.15 K$ $T_2 = 35^{\circ }C + 273.15 = 308.15 K$ Now we can plug in the values into the formula: $\frac{101.3 kPa}{298.15 K} = \frac{P_2}{308.15 K}$ Solving for $P_2$, we get: $P_2 = \frac{101.3 kPa \times 308.15 K}{298.15 K} \approx 104.6 kPa$ Therefore, the pressure change when the gas is heated from $25^{\circ }C$ to $35^{\circ }C$ is approximately 3.3 kPa. 2. To find the temperature at standard pressure for a gas with a pressure of 794 mm Hg at $43^{\circ }C$, we can use the same formula for Amonton's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ First, we need to convert the pressure from mm Hg to atm by dividing by 760: $P_1 = \frac{794 mm Hg}{760 mm Hg/atm} \approx 1.045 atm$ Now we can plug in the values into the formula: $\frac{1.045 atm}{316.15 K} = \frac{1 atm}{T_2}$ Solving for $T_2$, we get: $T_2 = \frac{1 atm \times 316.15 K}{1.045 atm} \approx 303.15 K$ Therefore, the temperature at standard pressure for the gas with a pressure of 794 mm Hg at $43^{\circ }C$ is approximately 303.15 K. 3. To find the final temperature of a gas when pressurized from 12.0 atm to 25.0 atm and its original temperature was $32.0^{\circ }C$, we can use the same formula for Amonton's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ First, we need to convert the temperature from Celsius to Kelvin by adding 273.15: $T_1 = 32.0^{\circ }C + 273.15 = 305.15 K$ Now we can plug in the values into the formula: $\frac{12.0 atm}{305.15 K} = \frac{25.0 atm}{T_2}$ Solving for $T_2$, we get: $T_2 = \frac{25.0 atm \times 305.15 K}{12.0 atm} \approx 635.29 K$ Therefore, the final temperature of the gas when pressurized from 12.0 atm to 25.0 atm and its original temperature was $32.0^{\circ }C$ is approximately 635.29 K. 4. To find the final pressure of a sample of gas at 3000 mm Hg inside a steel tank when cooled from $500^{\circ }C$ to $0^{\circ }$, we can use the same formula for Amonton's Law: $\frac{P_1}{T_1} = \frac{P_2}{T_2}$ First, we need to convert the temperatures from Celsius to Kelvin by adding 273.15: $T_1 = 500^{\circ }C + 273.15 = 773.15 K$ $T_2 = 0^{\circ }C + 273.15 = 273.15 K$ Now we can plug in the values into the formula: $\frac{3000 mm Hg}{773.15 K} = \frac{P_2}{273.15 K}$ Solving for $P_2$, we get: $P_2 = \frac{3000 mm Hg \times 273.15 K}{773.15 K} \approx 1100 mm Hg$

Problemas

Solución

Responder