Question #1.) A fixed volume of gas at constant temperature has a volume of 45.6

Question

Question #1.) A fixed volume of gas at constant temperature has a volume of 45.6 L at 1.24 atm. Use Boyle's Law to calculate:

a. the volume of the gas when compressed to 3.53 atm.

b. the pressure of the gas when it is compressed to 12.55 L.

Question #2.) An aerosol can with a volume od 445 mL contains 0.355 grams of propane (C3H8) as a propellant. What is the pressure inside the can at 27 degrees celsius?

Question #3.) At 46 degree celsius and 1.235 atm a gas occupies a volume of 0.735 L. How many moles of gas are present? What would be the pressure of the gas at 105 degrees celsius and a volume of 1.00L? How many liters would the gas occupy at STP?

Explanation / Answer

Answer – 1) We are given, V1 = 45.6 L , P1 = 1.24 atm

a) P2 = 3.3 atm , V2 = ? we need to use the Boyle’s law and we know formula for this law

P1V1 = P2V2

So, V2 = P1V1/P2

             = 45.6 L * 1.24 atm / 3.53 atm

             = 16.02 L

So, the volume of the gas when compressed to 3.53 atm is 16.0 L

b) V2 = 12.55 L

we know, P1V1 = P2V2

P2 = P1V1 / V2

     = 45.6 L * 1.24 atm / 12.55 L

     = 4.50 atm

So, the pressure of the gas when it is compressed to 12.55 L is 4.50 atm

2) We are given, V = 445 mL = 0.445 L

T = 27 +273 = 300 K , mass of propane = 0.355 g , P = ?

First we need to calculate moles of propane

Moles of propane = 0.355 g / 44.096 g.mol-1

                             = 0.00805 moles

Now we need to use Ideal gas law

PV = nRT

So, P = nRT/V

         = 0.00805 moles * 0.0821 L.atm.mol-1.K-1 * 300 K / 0.445 L

          = 0.446 atm

So, the pressure inside the can at 27 degrees Celsius is 0.446 atm

3) Given, T = 46+273 = 319 K , P = 1.235 atm , V = 0.735 L , so n = ?

We know Ideal gas law

PV = nRT

So, n = PV/RT

         = 1.235 atm * 0.735 L / 0.0821 L.atm.mol-1.K-1 * 319 K

          = 0.0347 moles

So, moles of gas are present is 0.0347 moles

Now we are given, T = 105 + 273 = 378 K , , V = 1.00 L we calculated, n= 0.0347

So, P = nRT/V

         = 0.0347 moles * 0.0821 L.atm.mol-1.K-1 * 378 K / 1.00 L

          = 1.08 atm

So, the pressure of the gas at 105 degrees celsius and a volume of 1.00L is 1.08 atm.

We know at STP, 1 mole = 22.4 L

So, 0.0347 moles = ?

= 0.776 L

So, 0.776 liters would the gas occupy at STP.

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