75L stainless steel container was charged with 3 atm of hydrogen gas and 4 atm of oxygen gas. a spark ignited the misture, producing water. what is the pressure of the tank at 25C? at 125C?
The 3 atm (partial pressure) of H2 will combine with 1.5 atm (partial pressure) of O2 to produce 3 atm (partial pressure) of H2O, with 2.5 atm (partial pressure) of O2 left over. Thus 7 atm of reactants produces 5.5 atm of products + unreacted O2.
I am assuming that the inital and final temperatures are what you have specified. This means that the heat of reaction will have been transfered out, as would be the case with a calorimeter. In that situation, with constant T and volume, the final pressure is proportional to the number of moles present, and that factor is 5.5/7 = 0.7857, at both temperatures.
I made a mistake in my previous answer by not considering the condensation of H2O. At 25C, almost all of the H20 produced will be in liquid form, so the actual pressure of the gas after cooling down the reaction products to 25 C will be 2.5/7 = 35.7% of the initial pressure. At 125 C, most (but not all) of the H2O produced by combustion will be in gaseous form. You will have to make use of data on the saturation pressure of H2O to get exact answers.
To determine the pressure of the tank after the reaction has occurred, we need to consider the ideal gas law. The ideal gas law equation is PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin.
First, let's find the number of moles of each gas. We can use the ideal gas law equation to calculate the number of moles.
For hydrogen gas:
PV = nRT
3 atm * 75 L = n * (0.0821 L * atm / mol * K) * (25 + 273) K
225 atm * L = n * 0.0821 L * atm / mol * K * 298 K
n (moles of H2) = (225 atm * L) / (0.0821 L * atm / mol * K * 298 K)
n (moles of H2) = 9.15 mol (rounded to two decimal places)
For oxygen gas:
PV = nRT
4 atm * 75 L = n * (0.0821 L * atm / mol * K) * (25 + 273) K
300 atm * L = n * 0.0821 L * atm / mol * K * 298 K
n (moles of O2) = (300 atm * L) / (0.0821 L * atm / mol * K * 298 K)
n (moles of O2) = 12.18 mol (rounded to two decimal places)
Now, let's calculate the total number of moles of gas in the tank:
Total moles of gas = moles of H2 + moles of O2
Total moles of gas = 9.15 mol + 12.18 mol
Total moles of gas = 21.33 mol (rounded to two decimal places)
To find the pressure of the tank after the reaction, we can use the ideal gas law equation with the new volume and temperature.
For 25°C (298 K):
PV = nRT
P * 75 L = 21.33 mol * (0.0821 L * atm / mol * K) * 298 K
P = (21.33 mol * 0.0821 L * atm / mol * K * 298 K) / 75 L
P ≈ 6.81 atm (rounded to two decimal places)
For 125°C (398 K):
PV = nRT
P * 75 L = 21.33 mol * (0.0821 L * atm / mol * K) * 398 K
P = (21.33 mol * 0.0821 L * atm / mol * K * 398 K) / 75 L
P ≈ 9.13 atm (rounded to two decimal places)
Therefore, the pressure of the tank at 25°C is approximately 6.81 atm, and at 125°C, it is approximately 9.13 atm.