Question

2. 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. T
(a) An additional 1.80 mol of argon gas is added to the container. [ans: 20.0 L]
(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. [ans: 5.00 L]
3. A balloon that contains 4.80 g of carbon dioxide gas has a volume of 12.0 L. Assume
that the pressure and temperature of the balloon remain constant. What is the new
volume of the balloon if an additional 0.50 mol of CO₂ is added? [ans: 67 L]

Answer 1

Hi goodmorning

The total volume of the gas is then 12.0 L + 0.806 L = 12.806 L.

To answer these questions, you can use the ideal gas law, which relates the pressure, volume, temperature, and number of moles of a gas. The ideal gas law is given by the following equation:

PV = nRT

Where P is the pressure of the gas, V is the volume of the gas, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature of the gas.

If pressure and temperature remain constant, then the volume of the gas will change inversely with the number of moles of the gas. This means that if the number of moles of the gas increases, the volume of the gas will decrease, and if the number of moles of the gas decreases, the volume of the gas will increase.

For example, in part (a), the number of moles of argon gas in the container increases by 1.80 mol, so the volume of the gas will decrease. The new volume of the gas can be calculated as follows:

V = (nRT)/P

= (1.80 mol * 8.31 J/mol*K * 300 K)/(1 atm)

= 4452 J

= 4.452 L

The total volume of the gas is then 10.00 L + 4.452 L = 14.452 L.

In part (b), the number of moles of argon gas in the container increases by a certain amount, which you can calculate using the molar mass of argon. The molar mass of argon is 39.948 g/mol, so the number of moles of argon in 25.0 g of argon is 25.0 g / 39.948 g/mol = 0.625 mol. The new volume of the gas can be calculated as follows:

V = (nRT)/P

= (0.625 mol * 8.31 J/mol*K * 300 K)/(1 atm)

= 1478.125 J

= 1.478 L

The total volume of the gas is then 10.00 L + 1.478 L = 11.478 L.

In part (c), the number of moles of argon gas in the container decreases by half, so the volume of the gas will increase. The new volume of the gas can be calculated as follows:

V = (nRT)/P

= (0.9 mol * 8.31 J/mol*K * 300 K)/(1 atm)

= 2463.9 J

= 2.464 L

The total volume of the gas is then 10.00 L + 2.464 L = 12.464 L.

In part (d), the number of moles of CO2 gas in the balloon increases by 0.50 mol, so the volume of the gas will decrease. The molar mass of CO2 is 44.01 g/mol, so the number of moles of CO2 in 4.80 g of CO2 is 4.80 g / 44.01 g/mol = 0.109 mol. The new volume of the balloon can be calculated as follows:

V = (nRT)/P

= (0.109 mol + 0.50 mol) * 8.31 J/mol*K * 300 K)/(1 atm)

= 806.36 J

= 0.806 L