I have some chemistry questions.?

[somemoonlitnite]

1. The volume of a gas at 93.0 kPa is 500.0 mL. If the pressure is increased to 221 kPa, what will be the new volume?

2. A gas at 99°C occupies a volume of 0.67 L. At what Celsius temperature will the volume increase to 1.39 L?

3. A sample of gas of unknown pressure occupies 0.779 L at a temperature of 294 K. The same sample of gas is then tested under known conditions and has a pressure of 30.7 kPa and occupies 0.659 L at 303 K. What was the original pressure of the gas?

4. Determine the kelvin temperature required for 0.0538 mol of gas to fill a balloon to 1.20 L under 0.980 atm pressure.

5. How many liters of propane gas (C3H8) will undergo complete combustion with 30.4 L of oxygen gas?

Thanks for any help!

[skyjaya]

1) For this problem, we must use Boyle's Law, where P1V1=P2V2 at the same temperature. Pressure and volume are inversely related, that is, increasing your pressure would decrease the volume. So keep that in mind to check your answer after you've calculated it.

Rearrange the eq., so that we can solve for the final volume(V2). V2=(P1*V1) /P2

So, if P1=93.0kPa, V1=500mL, and P2=221kPa, then:

V2= (93.0kPa)(500mL) / 221kPa = 210 mL of gas.

This follows Boyle's Law because the volume has been reduced due to inc. pressure.

2) This is a Charle's Law (V1T2=V2T1). Pressure is constant. Volume and temp. are directly relate, so as we inc. the temp., we increase the volume. V1=0.67. T1=99 C + 273 = 372K. V2=1.39 L T2 = unknown.

T2 = (V2*T1) / V1 = (1.39L * 372 C) / (0.67 L) = 771 K.

771 K - 273 K = 498 K.

The increased temperature results in the increased volume given as V2. So we are good here.

3) Here, we must use the combined gas law, which combines both Boyle's and Charle's laws. This is used when the amount of gas is constant, but P, T, and V change.

P1V1 / T1 = P2V2 / T2
P1 is unknown, V1=.779L, T1= 294K. P2=30.74kPa, V2=.659L, T2=303K

Solving for P1, the equation reads:
P1 = (P2V2T1) / (V2T1)
P1 = (30.7kPa * 0.779L * 294K) / (303 K * 0.659L)
Appropriately, our units cancel out and we are left with units in kPa, pressure units.

P1 = 35.2 kPa.

4) This is an ideal gas law (PV = nRT ) problem. It is used to describe a gas at a single set of conditions. When we use this law, pressures must be in atm, temps in Kelvin, and Volumes in Liters. Convert if needed.

Solving for T, T =PV/nR, where R is the gas constant 0.821 L*atm / mol*K.

T=(0.980 atm)(1.20L) / (0.0538 mol)(0.821 Latm/mol K)
all units except 1/1/K, which is just K cancel out.
So, T = 266 K.

5) A combustion reaction goes as follows for propane:
C3H8 + 5O2 -----> 3CO2 + 4H20

You can enter values for O2 into the ideal gas law, solve for the moles of O2. Then using stoichiomenty (1 mole of propane = 5 moles of O2), you can get moles of propane. Using this value in the ideal gas law, you can get the volume of propane.

However, there is an easier way. Use volumes in stoichiometry the way you would use moles.
Given 30.4 L of O2, with 5 moles of O2 per mole of propane, you get :
30.4 L O2 * (1 mole propane/ 5 moles O2) = 6.08 L propane.
This is the same answer that you get from the more complicated ideal gas law calculations.

I hope this helps you out.

[Dr.A]

1
93.0 x 500 = 221 V2
V2 =210.4 mL ( at constant temperature)
2.
T1 = 99 + 273 = 372 K
0.67 / 372 = 1.39 / T2
T2 = 771.8 K
771.8 - 273 = 498.8°C ( at constant volume)
3.
p = 30700 Pa / 101325 =0.303 atm
n = pV / RT = 0.303 x 0.659 / 0.0821 x 303 = 0.00803
original pressure = 0.00803 x 0.0821 x 294 / 0.779 L =0.249 atm
4.
T = pV /n R = 0.980 x 1.20 / 0.0538 x 0.0821 =266 K
5.
C3H8 +5 O2 >> 3CO2 + 4 H2O
the ratio is 1 : 5
liters of O2 = 30.4 x 5 = 152 L