Imagine that you have a 6.00 L gas tank and a 3.00 L gas tank. You need to fill one tank with oxygen and the other with acetylene to use in conjunction with your welding torch. If you fill the larger tank with oxygen to a pressure of 115 atm, to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? Assume ideal behavior for all gases.
The balanced chemical reaction is:
2C2H2 + 5O2 --> 4CO2 + 2H2O
The mole ratio of C2H2 to O2 is 2 : 5. Rearranging the Ideal Gas Law, we get:
n = PV / RT or
Let n1 = P1V1/RT (moles of C2H2)
and n2 = P2V2/RT (moles of O2
n1/n2 = P1V1/RT * RT/P2V2
Simplifying,
n1/n2 = P1V1 / P2V2
n1/n2 = (P1/P2)(V1/V2)
Substitute:
n1 = moles of C2H2 (acetylene) = 2
n2 = moles of O2 = 5
P1 = pressure of acetylene (unknown)
P2 = pressure of O2
V1 = volume of C2H2
V2 = volume of O2
Write the equation.
2C2H2 + 5O2 ==> 4CO2 + 2H2O
2.Use PV = nRT to determine n (# mols) O2 in the larger tank. No T is given; therefore, pick any T and use that T throughout the calculations.
3. Using the coefficients in the balanced equation, convert mols O2 to mols C2H2.
4. Now use PV = nRT to calculate P of the smaller tank for C2H2.
The T is have to use is 25C,and im trying to find the number of moles of oxygen first, so
PV= nRT
(115 atm)(6L)= n(0.08206)(25)
bUH IM NOT SURE IF I NEED TO CONVERT THE PRESSURE OR LITRES TO SOMETHING ELSE
P is in atmospheres. V is in liters. T is in Kelvin (Kelvin = 273 + C or 298 for 25 C). R is 0.08205 L*atm/mol*K.
OMg thank you soo much
Well, isn't this an explosive situation we have here? Oxygen and acetylene, a perfect recipe for blowing things up! So, let's get down to business.
Firstly, you need to know the number of moles of oxygen and acetylene you have in each tank to figure out the pressure ratio. Remember, moles are like those clingy friends; they always stick around.
For the oxygen tank, we can use the ideal gas law, PV = nRT. Here, we know the volume (6.00 L), the pressure (115 atm), and R (the ideal gas constant). If we assume room temperature and convert to SI units, we can find the number of moles of oxygen.
Now, for the acetylene tank, we have a volume of 3.00 L. But here's the plot twist: acetylene is actually a gas that is dissolved in liquid acetone (yeah, I know, weird, right?). So, to figure out the number of moles of acetylene, we need to consider the equilibrium pressure of dissolved acetylene in acetone.
To ensure that both gases run out at the same time, we need the same number of moles in both tanks. So, equate the number of moles of oxygen and acetylene, and solve for the pressure of acetylene.
But hold on, before you start playing with dangerous gases, you should probably double-check with a professional or consult your local friendly welding expert. Safety first, my friend!
Just remember, if things go bad, you can always join a circus as the next explosive performer. Just make sure to wear a funny hat and a red nose!
To determine the pressure needed to fill the acetylene tank, we can use Boyle's Law, which states that the product of pressure and volume is constant at constant temperature.
First, let's calculate the initial amount of oxygen in moles in the 6.00 L tank. Since we know the pressure (115 atm), we can use the Ideal Gas Law equation:
PV = nRT
Here, 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.
Rearranging the equation, we have:
n = PV / RT
Given that the pressure is 115 atm, the volume is 6.00 L, and the ideal gas constant R is approximately 0.0821 L·atm/(mol·K), we can substitute the values into the equation to find the number of moles of oxygen.
n(oxygen) = (115 atm) * (6.00 L) / (0.0821 L·atm/mol·K * T)
Next, let's calculate the initial amount of acetylene in moles in the 3.00 L tank. Since we want to run out of oxygen and acetylene at the same time, the ratio of moles between oxygen and acetylene should be equal, assuming ideal behavior for all gases.
Therefore, n(oxygen) / V(oxygen) = n(acetylene) / V(acetylene)
Simplifying the equation, we have:
n(acetylene) = (n(oxygen) / V(oxygen)) * V(acetylene)
Substituting the values we calculated earlier, with V(acetylene) = 3.00 L, we can find the number of moles of acetylene needed.
Now, to find the pressure of acetylene, we can rearrange the Ideal Gas Law equation:
P(acetylene) = n(acetylene) * R * T / V(acetylene)
Since we want to run out of both gases at the same time, we can set the pressures of oxygen and acetylene equal to each other:
P(oxygen) = P(acetylene)
Substituting the values into the equation, we have:
(115 atm) = (n(acetylene) * R * T) / (3.00 L)
Now, solve for P(acetylene):
P(acetylene) = (115 atm) * (3.00 L) / (n(acetylene) * R * T)
By substituting the calculated values for n(acetylene) and all the known values, you can determine the pressure needed to fill the acetylene tank.