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.)
Supposing all the "gas" is water vapor.
The source below gives the specific heat of water vapor and liquid water:
(2.000 J/(g·°C) x ((0.07796 g) x (125 ºC - 100 ºC) = 3.898 J from cooling the vapor to 100 ºC
(0.07796 g H2O) / (18.01532 g H2O/mol) = 0.00432743 mol H2O
(40.7 kJ/mol) x (0.00432743 mol) = 0.1761 kJ = 176.1 J from condensing the vapor
(4.184 J/(g·°C)) x (0.07796 g H2O) x (100 ºC - 59.0 ºC) = 13.37 J from cooling the liquid water to 59.0 ºC
3.898 J + 176.1 J + 13.37 J = 193.4 J total
Well, this seems like a real gas-tly problem! But don't worry, I'll try to make it bearable.
To calculate the amount of heat released, we need to determine the change in enthalpy during the cooling process.
First, we'll calculate the initial number of moles of the gas using the ideal gas law. The given pressure, volume, and temperature are all at the same time point, so no extra calculations are required.
PV = nRT
Using the given values and converting Celsius to Kelvin, we have:
(1140 mmHg)(108 mL) = n(0.0821 L.atm/mol.K)(183 + 273.15 K)
Now, let's convert the pressure and volume to more standard units:
(1 atm/760 mmHg)(108 mL) = n(0.0821 L.atm/mol.K)(183 + 273.15 K)
0.142105 L = n(0.0821 L.atm/mol.K)(456.15 K)
n ≈ 0.0058 mol
Now that we know the initial number of moles, we can move on to calculating the amount of heat released during the cooling process. To do that, we'll first need to calculate the change in enthalpy during the phase changes of the water.
Since the given heat of vaporization is at 100 ºC and we're only going to 59 ºC, we'll need to consider the heat of vaporization at 100 ºC and the heat of condensation at 59 ºC. So, let's break it down:
Heat released during cooling from 125 ºC to 100 ºC:
ΔH1 = (n × heat of vaporization at 100 ºC)
Heat released during condensation from 100 ºC to 59 ºC:
ΔH2 = -(n × heat of vaporization at 100 ºC)
Heat released during cooling from 59 ºC to 0 ºC:
ΔH3 = (n × heat of fusion at 0 ºC)
Adding up all the changes:
ΔHtotal = ΔH1 + ΔH2 + ΔH3
Remember, when the water condenses, it releases heat, so that's why we have a negative sign for ΔH2.
Now, we can substitute the values and calculate the heat:
ΔHtotal = (0.0058 mol)(40.7 kJ/mol) + (0.0058 mol)(-40.7 kJ/mol) + (0.0058 mol)(-6.02 kJ/mol)
And voila! You just need to do the math to find the amount of heat released by the system. Just remember to keep track of the units in your calculations.
I hope this calculation has warmed up your brain a bit!
To calculate the amount of heat released by the system, we first need to determine the amount of gas that will transform into liquid water.
Step 1: Calculate the number of moles of gas.
We can use the ideal gas law equation to calculate the number of moles of gas.
PV = nRT
Where:
P = pressure = 1140 mmHg (which is equivalent to 1520 dyne/cm^2 or 1520 N/m^2)
V = volume = 108 mL (which is equivalent to 0.108 L)
n = number of moles (unknown)
R = ideal gas constant = 0.0821 L·atm/(K·mol)
T = temperature = 183 °C (which is equivalent to 456 K)
Rearranging the equation to solve for n:
n = (PV) / (RT)
n = (1520 N/m^2 * 0.108 L) / (0.0821 L·atm/(K·mol) * 456 K)
n = 83.72 mol
Step 2: Calculate the heat released from the gas cooling to 125 ºC.
We'll use the equation:
q = n * ΔHvap
Where:
q = heat released
n = number of moles of gas
ΔHvap = heat of vaporization of water = 40.7 kJ/mol
q = 83.72 mol * 40.7 kJ/mol
q = 3409.004 kJ
Step 3: Calculate the heat released from the gas condensing to liquid water at 59.0 ºC.
We'll use the equation:
q = n * ΔHfus
Where:
q = heat released
n = number of moles of gas
ΔHfus = heat of fusion of water = 6.02 kJ/mol
q = 83.72 mol * 6.02 kJ/mol
q = 503.0624 kJ
Step 4: Calculate the total heat released by adding the heat from both steps.
Total heat released = q1 + q2
Total heat released = 3409.004 kJ + 503.0624 kJ
Total heat released = 3912.0664 kJ
Therefore, the amount of heat released by the system is approximately 3912.0664 kJ.
To calculate the amount of heat released by the system, we need to consider two separate processes:
1) Cooling the gas sample from 183 ºC to 125 ºC.
2) Condensing the gas into liquid water at 59.0 ºC.
Let's calculate the heat released in each step:
1) Cooling the gas sample from 183 ºC to 125 ºC:
First, we need to calculate the change in temperature. ΔT = Tfinal - Tinitial = 125 ºC - 183 ºC = -58 ºC.
Next, we need to convert the amount of gas from milliliters to moles. We can use the ideal gas law 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.
Since the system is completely insulated, we can assume constant volume (V = 108 mL). Thus, we have:
n = (PV) / (RT)
To use this equation, we need to convert the pressure from mmHg to atm and the temperature from ºC to Kelvin:
P = 1140 mmHg * (1 atm / 760 mmHg) = 1.5 atm
T = 183 ºC + 273.15 = 456.15 K
Now, we can substitute these values into the equation:
n = (1.5 * 0.108) / (0.0821 * 456.15)
Using the equation, we find that n ≈ 0.0049 moles of gas.
To calculate the heat released during this process, we use the equation:
q = ΔT * n * C
Where q is the heat released, ΔT is the change in temperature, n is the number of moles, and C is the molar heat capacity of the gas.
Since the gas is assumed to be ideal, its molar heat capacity at constant volume (CV) is equal to (3/2)R, where R is the ideal gas constant. So:
C = (3/2) * R
Now we can calculate the heat released during this step:
q1 = (-58 ºC) * (0.0049 moles) * [(3/2) * (0.0821 atm L / mol K)]
2) Condensing the gas into liquid water at 59.0 ºC:
During this phase change, the gas releases heat equal to the enthalpy of vaporization of water.
The amount of heat released can be calculated using the equation: q2 = n * ΔHvap
Where q2 is the heat released, n is the number of moles, and ΔHvap is the enthalpy of vaporization.
Since we want to transform all the moles of gas, we can use the value of n from the previous calculations.
Now we can calculate the heat released during this step:
q2 = 0.0049 moles * (-40.7 kJ / mol)
Finally, to find the total heat released by the system, we sum up the values of q1 and q2:
q_total = q1 + q2
Remember to include the correct units in your final answer.