Answer:
Step-by-step explanation:
(a) When an additional 1.80 mol of argon gas is added to the container, the total number of moles of gas increases to 1.80 + 1.80 = 3.60 moles. Since the pressure and temperature of the gas remain constant, the volume of the gas will also remain constant at 10.00 L. So the new volume of the gas will be 10.00 L.
(b) When 25.0 g of argon gas is added to the container, we need to convert the mass of the gas to moles to determine its effect on the volume. We can use the ideal gas law to do this:
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 pressure and temperature remain constant, we can rearrange the equation to solve for n:
n = PV/RT
Plugging in the values, we get:
n = (1.00 atm)(25.0 g)/(0.08206 Latm/molK)(298 K)
This gives us a value of n = 0.34 mol.
Now that we know the number of moles of gas, we can use the ideal gas law again to solve for the new volume of the gas:
V = nRT/P
Plugging in the values, we get:
V = (0.34 mol)(0.08206 Latm/molK)(298 K)/(1.00 atm)
This gives us a new volume of V = 13.5 L.
(c) When half of the gas escapes from the container, the total number of moles of gas decreases to 1.80/2 = 0.90 mol. Again, we can use the ideal gas law to solve for the new volume of the gas:
V = nRT/P
Plugging in the values, we get:
V = (0.90 mol)(0.08206 Latm/molK)(298 K)/(1.00 atm)
This gives us a new volume of V = 7.50 L.
I hope this helps! Let me know if you have any questions