A 108 mL gas sample has a mass of 77.96 mg at a pressure of 1140 mmHg and a temperature of 183 °C. Assume ideal gas behavior and a completely insulated system, and please don’t forget units with your answers.
The system is cooled. Calculate the amount of heat that will be released by the system if all of the gas at 125 ºC transforms into liquid water at 59.0 ºC.
(The heat of vaporization of water at 100.0 ºC is 40.7 kJ/mol, and the heat of fusion of water at 0.0 ºC is 6.02 kJ/mol.)
First, let's calculate the number of moles of gas in the initial sample using the ideal gas equation:
PV = nRT
where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.
Since the pressure is given in mmHg, we need to convert it to atm:
1 atm = 760 mmHg
So the pressure in atm is 1140 mmHg / 760 mmHg = 1.5 atm.
Also, the temperature is given in °C, so we need to convert it to Kelvin:
T (K) = T (°C) + 273.15
Initial number of moles, n1 = (1.5 atm * 0.108 L) / (0.0821 atm L/mol K * (183 + 273.15) K)
n1 = 0.009823 mol
Now let's calculate the change in heat when the gas transforms into liquid water.
First, the gas needs to cool from 125 ºC to 59.0 ºC, so the heat lost during this process can be calculated using the equation:
q = mcΔT
where q is the heat lost, m is the mass of the gas, c is the specific heat capacity, and ΔT is the change in temperature.
The specific heat capacity of a gas at constant pressure (Cp) is approximately 29.1 J/mol K.
Heat lost during cooling, q1 = (0.07796 g / 0.009823 mol) * (29.1 J/mol K) * (125 °C - 59.0 °C)
q1 = 28,598 J
Next, the gas needs to condense into liquid water at 59.0 ºC.
The heat released during condensation can be calculated using the equation:
q = nΔHvap
where q is the heat released, n is the number of moles, and ΔHvap is the heat of vaporization.
Heat released during condensation, q2 = 0.009823 mol * (40.7 kJ/mol * 1000 J/kJ)
q2 = 398.4 J
Finally, the liquid water needs to cool from 59.0 ºC to 0.0 ºC.
The heat lost during this process can be calculated using the equation:
q = mcΔT
where q is the heat lost, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature.
The specific heat capacity of water is approximately 4.18 J/g°C.
Heat lost during cooling, q3 = (0.07796 g) * (4.18 J/g°C) * (59.0 °C - 0.0 °C)
q3 = 18,203 J
Therefore, the total amount of heat released by the system is:
q_total = q1 + q2 + q3
q_total = 28,598 J + 398.4 J + 18,203 J
q_total = 47,199 J
To calculate the heat released by the system, we need to consider the energy changes during phase transitions.
First, we need to find the number of moles of gas in the 108 mL sample. We can use the ideal gas equation to do this:
PV = nRT
Where:
P = pressure (in atm),
V = volume (in liters),
n = number of moles,
R = ideal gas constant (0.0821 L·atm/(mol·K)),
T = temperature (in Kelvin).
Converting the given pressure, volume, and temperature to the appropriate units:
Pressure: 1140 mmHg * (1 atm / 760 mmHg) = 1.5 atm
Volume: 108 mL * (1 L / 1000 mL) = 0.108 L
Temperature: 183 °C + 273.15 = 456.15 K
Now we can substitute these values into the ideal gas equation:
(1.5 atm)(0.108 L) = n(0.0821 L·atm/(mol·K))(456.15 K)
n = (1.5 atm * 0.108 L) / (0.0821 L·atm/(mol·K) * 456.15 K)
n = 0.00211 mol
Since we have determined the number of moles of gas, we can now calculate the heat released during the phase transitions.
Step 1: Gas at 125 ºC to gas at 100 ºC
For this step, we don't need to consider any phase changes because the gas remains in its gaseous state. We only need to consider the change in temperature.
The heat released for this step can be calculated using the formula:
q = n*Cp*ΔT
Where:
q = heat released,
n = number of moles,
Cp = molar heat capacity (in J/mol·K),
ΔT = change in temperature (in K).
The molar heat capacity of an ideal gas at constant pressure is approximately 29.1 J/(mol·K).
ΔT = (125 ºC + 273.15 K) - (100 ºC + 273.15 K) = 25 K
q = (0.00211 mol)(29.1 J/(mol·K))(25 K)
q = 1.496 J
Step 2: Gas at 100 ºC to liquid water at 100 ºC
For this step, the gas undergoes a phase change from gas to liquid water at the boiling point.
The heat released during this step can be calculated using the formula:
q = n*ΔHvap
Where:
q = heat released,
n = number of moles,
ΔHvap = molar heat of vaporization (in J/mol).
The molar heat of vaporization for water at 100 ºC is given as 40.7 kJ/mol. Let's convert it to J/mol:
ΔHvap = 40.7 kJ/mol * (1000 J / 1 kJ)
ΔHvap = 40,700 J/mol
q = (0.00211 mol)(40,700 J/mol)
q = 85.987 J
Step 3: Liquid water at 100 ºC to liquid water at 59.0 ºC
For this step, the liquid water undergoes a change in temperature.
The heat released during this step can be calculated using the formula:
q = n*Cp*ΔT
Where:
q = heat released,
n = number of moles,
Cp = molar heat capacity (in J/mol·K),
ΔT = change in temperature (in K).
The molar heat capacity for liquid water is approximately 75.3 J/(mol·K).
ΔT = (100 ºC + 273.15 K) - (59.0 ºC + 273.15 K) = 13.85 K
q = (0.00211 mol)(75.3 J/(mol·K))(13.85 K)
q = 2.477 J
To find the total heat released by the system, we sum the heat released in each step:
Total heat released = q1 + q2 + q3 = 1.496 J + 85.987 J + 2.477 J = 90.96 J
Therefore, the amount of heat released by the system during the phase change is 90.96 J.