**Gas Practice Questions****1.) **Convert
539 torr to atm.

**2.) **A gas takes up 25.2 liters at 25^{o}C. At 25^{o}C,
the gas can also take up 12.2 liters at 1500 torr. What was the pressure, in atm, of the
original sample?

**3.) **A gas takes up 14.8 liters of 24^{o}C. What temperature in
kelvin is required to obtain a volume of 25.0 liters at constant pressure?

**4.) **How many moles of chlorine gas are present at 25^{o}C, 762
torr, with a volume of 14.2 L?

**5.) **If "Bor" (thats the name of a person) has a sample of
gas that has a volume of 8.2 liter at 25^{o}C and 2 atm, how much volume will it
take up if you decrease pressure to 1.5 atms and icrease temperature to 100^{o}C?

**6.) **25 liters of gas A is pumped into a container at 25^{o}C
and 760 torr with 20 liters of gas B at 25^{o}C and 700 torr. Calculate the total
pressure when both gases are pumped into a tank with 10 liters at 25^{o}C.

**7.) **Calculate the mole fraction of oxygen when 200 torr of air (760
torr total) is oxygen.

**8.) **KClO_{3} is decomposed by the following reaction:

2KClO_{3}(s) ---> 2KCl(s) + 3O_{2}(g)

The O_{2} produced was collected by the displacement of water at 22^{o}C
at a total pressure of 760 torr. The volume of gas collected was 1.20 liters, and the
vapor pressure of water at 22^{o}C is 21 torr. Calculate the partial pressure of O_{2}
in the gas collected and the mass of KClO_{3} in the sample that was decomposed.

**9.) **Calculate the root mean square velocity of hydrogen gas at 25^{o}C.

**10.) **What is the ratio of effusion between fluorine and chlorine?

**1.) **Just set up a simple ratio.

539 torr * (1 atm/760 torr) =** 0.709 atm**

**2.) **Use this formula:

P_{1}V_{1} = P_{2}V_{2}

460 atm * 1.20 liters = 839 torr * V_{2}

V_{2}= **0.658 L**

**3.) **You will use the ratio P_{1}V_{1} = P_{2}V_{2}
in order to solve this equation.

P_{1}V_{1} = P_{2}V_{2}

25.2 liters * P_{1} = 12.2 liters * (1500 torr / 760 torr)

Note that you have to convert the torr to atm. The problem asks for the final pressure
in atms. You could of just solved the ratio without the conversion, and converted the
answer later. Either way is acceptable.

25.2 liters * P_{1} = 24

**P**_{1} = 0.956 atm

**4.) **Use the formula:

V_{1}/T_{1} = V_{2}/T_{2}

14.8 liters / 297 torr = 25.0 liters / x

14.8 x = 7425

x = **502 K **

**5.) **PV = nRT (the ideal gas law! Note: you will see this a LOT in
gases, so memorize it)

n = PV/RT

P = (762 torr/760 torr) = 1.00atm

n = (1.00 atm * 14.2 L) / (298^{o}K * 0.08206)

n = **0.582 moles**

**6.) **PV = nRT

n = PV/RT

= (2 atm * 8.2 L) / (0.08206 * 298^{o}K) = 0.671 moles

Now, you must remember that you are taking this amount and changing it some more. So,
you do this again, this time, solving for VOLUME. (No, the initial 8.2 doesn't count,
because you are manipulating the environment)

PV = nRT

V = nRT/P

=(0.671 moles * 0.08206 * 373^{o}K) / 1.5 atm = **13.76 liters**

**7.) **You must solve this twice for both gas A and B, then combine the
individual pressures to obtain the final pressure. Use the Ideal Gas Law.

n_{gas a} = PV/RT = (1 atm * 25 liters) / (0.08206 * 298 K) = 1.02 moles

n_{gas b} = PV/RT = (1 atm * 20 liters) / (0.08206 * 298) = 0.818 moles

You know how many moles you have. The second part of the problem calls for a
manipulation of the environment, which means that you will be solving for PRESSURE.

P_{gas a} = nRT/V = (1.02 moles * 0.08206 * 298K / 10.0 liters) = 2.49 atm

P_{gas b} = nRT/V = (0.818 moles * 0.08206 * 298 K / 10.0 liters) = 2.00 atm

2.49 atm + 2.00 atm = **4.49 atm**

**8.) **This is a simple mole fraction. You know that air has a total of
760 torr. Since we are given that O_{2} directly contributes 200 of that, we can
set up this ratio:

O_{2} pressure/ total pressure = 200 torr / 760 torr = 0.263 = **26.3% **

**9.) **Remember that:

P_{total} = P_{O2} + P_{H2O} = P_{O2}
+ 21 torr = 760 torr

P_{O2} = **739 torr**

PV = nRT or n=PV/RT

P = 739 torr/760 torr = 0.972 atm

n = (0.972 atm * 1.20 liters) / (0.08206 * 295 K) = 0.0482 moles O_{2}

The final answer calls for mass of the KClO_{3}, not the O2. So we must find
the number of moles of KClO_{3}.

0.0482 moles O_{2} * (2 mol KClO_{3}/3 mol O_{2}) = 0.0321
moles KClO_{3}

0.0321 mol KClO_{3} * (122.6 g KClO_{3} / 1 mol KClO_{3}) = **3.94
grams KClO**_{3}

**10.) **Remember that u_{rms} formula is:

So, using that, let's fill in our variables and constants. R, for this equation is
8.3145 J/K*mol and T = 298 K. Calculating for M =

M_{H2} = 1 g/mol * 2 mol (since H_{2} is diatomic) * 1
kg/1000g = 0.002 kg

(3 * 8.3145 * 298/ 0.002) = 3716581.5 (you have to take the square root of this)
-----> **1928 m/s**

**11.) **Both fluorine and chlorine are diatomic. The equation relating
effusion and diffusion is:

Taking this, you take the square root of 70.90 g (which is the weight of Cl_{2}),
and divide it by the square root of 37.996 g (which is the weight of F_{2}). The
answer is 1.366. This means that fluorine effuses 1.366 times as fast as chlorine, which
makes sense because F_{2} is smaller.