At 35.0°c and 3.00 atm pressure, a gas has a volume of 1.40 l. what pressure does the gas have at 0.00°c and a volume of 0.950 l? which equation should you use? a balloon containing 0.500 mol ar at 0.00°c and 65.0 kpa pressure is expanded by adding more argon. how many moles of argon are added to bring the sample to a final volume of 60.0 l at 30.0°c and 45.0 kpa? what is the original volume of the gas? l

Answer :

The final pressure of gas will be, 3.92 atm

The original volume of gas is, 17.46 L

The number of moles of argon gas added is, 0.57 mol.

Explanation :

Part 1 :

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

where,

= initial pressure of gas = 3.00 atm

= final pressure of gas = ?

= initial volume of gas = 1.40 L

= final volume of gas = 0.950 L

= initial temperature of gas =

= final temperature of gas =

Now put all the given values in the above equation, we get:

Therefore, the final pressure of gas will be, 3.92 atm

Part 2 :

First we have to calculate the original volume of gas.

Using ideal gas equation:

where,

P = pressure of gas = 65.0 kPa

V = volume of gas = 3.06 L

T = temperature of gas =

n = number of moles of gas = 0.500 mol

R = gas constant   = 8.314 kPa.L/mol.K

Now put all the given values in the ideal gas equation, we get:

Now we have to calculate the final moles of sample of gas.

Using ideal gas equation:

where,

P = pressure of gas = 45.0 kPa

V = volume of gas = 60.0 L

T = temperature of gas =

n = number of moles of gas = ?

R = gas constant   = 8.314 kPa.L/mol.K

Now put all the given values in the ideal gas equation, we get:

Now we have to calculate the number of moles of argon gas added.

Moles of argon gas added = Final moles of gas - Initial moles of gas

Moles of argon gas added = 1.07 mol - 0.500 mol

Moles of argon gas added = 0.57 mol

Thus, the number of moles of argon gas added is, 0.57 mol.

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