Question

Please answer this with work shown for my chemistry homework please!!!

A sample containing 1.80 mol of argon gas has a volume of 10.00 L. What is
the new volume of the gas, in litres, when each of the following changes occurs in
the quantity of the gas? Assume that pressure and temperature remain constant.
The changes are not cumulative. ™
(a) An additional 1.80 mol of argon gas is added to the container. [ans: 20.0 LJ
(b) A sample of 25.0 g of argon gas is added to the container. [ans: 13.5 LJ
(c) A hole in the container allows half of the gas to escape.

Answer 1

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Step-by-step explanation:

(a) An additional 1.80 mol of argon gas is added to the container.

The new volume of the gas would be 20.0 L.

(b) A sample of 25.0 g of argon gas is added to the container.

To determine the new volume of the gas, we need to first determine how many moles of argon gas are present in the 25.0 g sample. To do this, we can use the molar mass of argon, which is 39.948 g/mol. This gives us a total of 0.625 moles of argon in the 25.0 g sample. Since the pressure and temperature are constant, the volume of the gas will increase in proportion to the number of moles of gas present. Since 0.625 moles of gas is being added to 1.80 moles of gas, the total number of moles of gas will be 2.425 moles. The new volume of the gas will be 10.00 L * (2.425 moles / 1.80 moles) = 13.5 L.

(c) A hole in the container allows half of the gas to escape.

The new volume of the gas would be 10.00 L / 2 = 5.00 L.

Answer 2

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