Problems #1 - 15

The form of the Combined Gas Law most often used is this:

(P_{1}V_{1}) / T_{1}= (P_{2}V_{2}) / T_{2}

Most commonly V_{2} is being solved for. The rearrangement looks like this:

V_{2}= (P_{1}V_{1}T_{2}) / (T_{1}P_{2})

A reminder: all these problems use Kelvin for the temperature. I will not usually comment on the change from °C to K. I will use 273 but be aware that your teacher (or computer lesson) may insist on using 273.15.

When you use the combined gas law paired with Dalton's Law, remember that a gas collected over water is always considered to be saturated with water vapor. The vapor pressure of water varies with temperature and must be looked up in a reference source.

**Problem #1:** A gas has a volume of 800.0 mL at −23.0 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure?

**Solution:**

1) Set up all the problem values in a solution matrix:

P _{1}= 300.0 torrP _{2}= 600.0 torrV _{1}= 800.0 mLV _{2}= xT _{1}= 250. KT _{2}= 500. K

2) The combined gas law is rearranged to isolate V_{2}:

P _{1}V_{1}T_{2}V _{2}= –––––– P _{2}T_{1}

3) Values are inserted into the proper places:

(300.0 torr) (800.0 mL) (500.0 K) V _{2}= ––––––––––––––––––––––––––– (250.0 K) (600.0 torr) V

_{2}= 800.0 mL

**Problem #2:** 500.0 liters of a gas in a flexible-walled container are prepared at 700.0 mmHg and 200.0 °C. The gas is placed into a tank under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the volume of the gas?

**Solution:**

1) The combined gas law is rearranged to isolate V_{2}:

V_{2}= (P_{1}V_{1}T_{2}) / (T_{1}P_{2})

2) Values are inserted into the proper places:

V_{2}= [(0.92105) (500) (293)] / [(473) (30)]V

_{2}= 9.51 L

Note the use of square brackets to communicate the correct order of operations.

Note that the problem provides different pressure units for the starting and ending values. I used 700/760 to convert from mmHg to atm.

Note that I paid scant attention to setting up the problem with correct sig figs in the problem. This happens often in gas law problems. Note also I omitted all the units.

**Problem #3:** 690.0 mL of oxygen are collected over water at 26.0 °C and a total pressure of 725.0 mm of mercury. What is the volume of dry oxygen at 52.0 °C and 800.0 mm pressure?

**Solution:**

1) Use Dalton's Law to remove the pressure of the water vapor:

P_{total}= P_{O2}+ P_{H2O}P

_{O2}= P_{total}- P_{H2O}P

_{O2}= 725.0 mmHg - 25.2 mmHg = 699.8 mmHgThe 25.2 value came from here. I looked up the value associated with 26.0 °C and converted it from kPa to mmHg following the instructions given.

2) Here are the values in a solution matrix:

P _{1}= 699.8 mmHgP _{2}= 800.0 mmHgV _{1}= 690.0 mLV _{2}= xT _{1}= 299.0 KT _{2}= 325.0 KA common student error is to use Dalton's Law, but then use the total pressure value in the combined gas law instead of using the correct value.

The correct pressure to use for P

_{1}is the 699.8 value, not the 725 value. The 725 is the pressure of an oxygen/water mixture and we want ONLY the oxygen (which is the 699.8 value).

3) Use the combined gas law:

x = [(699.8) (690.0) (325)] / [(299) (800.0)]x = 656 mL (to three sig figs)

**Problem #4:** What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300.0 L at 0.250 atm and 400.0 K.

**Solution:**

1) Here are the values in a solution matrix:

P _{1}= 0.250 atmP _{2}= 2.00 atmV _{1}= 300.0 LV _{2}= xT _{1}= 400.0 KT _{2}= 200.0 KNote how the problem statement is worded so as to give the starting values last.

2) The combined gas law rearranged to isolate V_{2}:

V_{2}= (P_{1}V_{1}T_{2}) / (T_{1}P_{2})x = [(0.25) (300) (200)] / [(400) (2)] <--- note lack of units

x = 18.75 L

To three sig figs, this is 18.8 L

**Problem #5:** At conditions of 785.0 torr of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 745.0 torr and 30.0 °C?

**Solution:**

V_{2}= [(785 mmHg) (45.5 mL) (303 K)] / [(288 K) (745 mmHg)]V

_{2}= 50.3757 mLTo three sig figs, the answer is 50.4 mL

**Problem #6:** What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 °C to 30.0 °C and a pressure change from standard pressure to 360.0 mmHg?

**Solution:**

We are looking to determine V_{2} in this problem. Here's the set up:

P _{1}= 760.0 mmHgP _{2}= 360.0 mmHgV _{1}= 400.0 LV _{2}= xT _{1}= 295 KT _{2}= 303 KV

_{2}= [(760 mmHg) (400 mL) (303 K)] / [(295 K) (360 mmHg)]V

_{2}= 867 mL (to three sig figs)

**Problem #7:** 400.0 mL of hydrogen are collected over water at 18.0 °C and a total pressure of 740.0 mm of mercury.

(a) What is the partial pressure of H_{2}?

(b) What is the partial pressure of H_{2}O?

(c) What is the volume of DRY hydrogen at STP?

**Solution:**

1) We will use Dalton's Law to determine the partial pressure of the dry hydrogen gas. We look up the vapor pressure of water in a reference source.

P_{total}= P_{H2}+ P_{H2O}P

_{H2}= P_{total}- P_{H2O}740.0 - 15.5 = 724.5 mmHg

I used a different reference source than previously used for the vapor pressure of water. There are many available on the Internet.

2) The partial pressure of the water is its vapor pressure of 15.5 mmHg.

3) Combined gas law rearranged to show V_{2} isolated:

P _{1}V_{1}T_{2}V _{2}= –––––– P _{2}T_{1}

(724.5 mmHg) (400.0 mL) (273 K) V _{2}= –––––––––––––––––––––––––––– (291 K) (760.0 mmHg) V

_{2}= 358 mL (to three sig figs)

**Problem #8:** The pressure of a gas is reduced to 75% of its initial value and the volume is increased by 40% of its initial value. Find the final temperature, given that the initial temperature was −10 °C.

**Solution:**

Let us assign P_{1}= 1, therefore P_{2}= 0.75

Let us assign V_{1}= 1, therefore V_{2}= 1.4I won't bother with units on P or V. Your teacher may want the units added in, so I'll do that below.

T

_{1}= −10 °C = 263 KP

_{1}V_{1}/T_{1}= P_{2}V_{2}/T_{2}[(1 atm) (1 L)] / 263 K = [(0.75 atm) (1.4 L)] / x

(1 atm) (1 L) (x) = (263 K) (0.75 atm) (1.4 L)

x = 276.15 K = 3.15 °C

**Problem #9:** The pressure of 8.06 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?

**Solution:**

1) Assign values as follows:

P _{1}= 3.00 atmP _{2}= 1.00 atmV _{1}= 8.06 LV _{2}= xT _{1}= 2.00 KT _{2}= 1.00 KNote the made up values for P and T.

2) Insert values into the combined gas law equation and solve for x:

P_{1}V_{1}/ T_{1}= P_{2}V_{2}/ T_{2}[(3.00 atm) (8.06 L)] / 2.00 K = [(1.00 atm) (x)] / 1.00 K

x = 12.1 L (to three sig figs)

**Problem #10:** A balloon of air now occupies 10.0 L at 25.0 °C and 1.00 atm. What temperature was it initially, if it occupied 9.40 L and was in a freezer with a pressure of 0.939 atm?

**Solution:**

1) Assign values as follows:

P _{1}= 0.939 atmP _{2}= 1.00 atmV _{1}= 9.40 LV _{2}= 10.0 LT _{1}= xT _{2}= 298 KNote how the problem gives you the ending conditions regarding PVT and asks you for a starting condition. Also, note that temperature is asked for. Compare this to the usual case of asking for the final volume.

2) Let's rearrange the combined gas law equation for T_{1}:

P _{1}V_{1}P _{2}V_{2}––––– = ––––– T _{1}T _{2}T

_{1}P_{2}V_{2}= P_{1}V_{1}T_{2}

P _{1}V_{1}T_{2}T _{1}= –––––– P _{2}V_{2}

3) Put values in and solve:

(0.939 atm) (9.40 L) (298 K) T _{1}= –––––––––––––––––––––––– (1.00 atm) (10.0 L) T

_{1}= 263 KThe form of the final temperature was not specificed, but it is usually asked for in Celsius, so:

T

_{1}= −10. °C

**Problem #11:** A gas occupies an 8.00 mL flexible-walled container. The pressure is doubled, the absolute temperature is quadrupled, and 15% of the gas leaks out. What is the new volume?

**Solution:**

1) This problem involves the combined gas law that has all four variables in it:

P _{1}V_{1}P _{2}V_{2}––––– = ––––– n _{1}T_{1}n _{2}T_{2}

2) The changes are all expressed in a relative way, so I will assume 1.00 atm and 1.00 K:

P _{1}= 1.00 atmP _{2}= 2.00 atmV _{1}= 8.00 mLV _{2}= xn _{1}= 1.00 moln _{2}= 0.85 molT _{1}= 1.00 KT _{2}= 4.00 KBy the way, having a gas at 1.00 K is pretty much an impossible thing. Pure helium-3 liquifies at about 3.2 K. No gas exists as a gas at 1 K. The point, of course, is to make the absolute temperature quadruple. We could use 100 K and 400 K and get the same answer. Or use 200 K and 800 K. The key point is that the temperature quadruples. And, note that it is the absolute temperature (in K) that quadruples, not the temperature in degrees Celsius.

Notice how I interpreted the 15% to mean 15% of the moles of gas escaped. If I decided that 15% of the mass escaped, the problem answer would be the same. I will leave it to you to figure out, if you so desire.

3) I won't bother to isolate V_{2} this time:

(1.00 atm) (8.00 mL) (2.00 atm) (x) ––––––––––––––––– = ––––––––––––––– (1.00 mol) (1.00 K) (0.85 mol) (4.00 K)

4) Cross-multiply:

(1.00 atm) (8.00 mL) (0.85 mol) (4.00 K) = (2.00 atm) (x) (1.00 mol) (1.00 K)

4) Divide:

(1.00 atm) (8.00 mL) (0.85 mol) (4.00 K) x = ––––––––––––––––––––––––––––––––– (2.00 atm) (1.00 mol) (1.00 K) x = 13.6 mL

**Problem #12:** What is the density, in g/L, for a gaseous compound at STP if the gas in a 1.00 L bulb weighs 0.672 g at 25.0 °C and 733.4 mm Hg?

**Solution:**

1) We need to know the volume of the gas at STP. For this, we use the combined gas law. Here's the data:

P _{1}= 733.4 mmHgP _{2}= 760.0 mmHgV _{1}= 1.00 LV _{2}= xT _{1}= 298 KT _{2}= 273 K

2) Here's the combined gas law with the data filled in:

(733.4 mmHg) (1.00 L) (760.0 mmHg) (x) ––––––––––––––––––– = ––––––––––––––– 298 K 273 K After a bit of math, we find:

x = 0.884 L

3) We are now ready to determine the density:

0.672 g / 0.884 L = 0.760 g/L

**Problem #13:** Predict how the volume of a given mass of gas will differ when the following changes in the temperature and pressure are made:

(a) The pressure is tripled while the absolute temperature is doubled.

(b) The absolute temperature is doubled while the pressure is reduced by half.

(c) The pressure and the absolute temperature are both doubled.

(d) The temperature is increased by four times while at the same time the pressure is doubled.

**Solution to (a):**

1) Place "fake" values into a solution matrix:

P _{1}= 1 kPaP _{2}= 3 kPaV _{1}= 1 LV _{2}= ???T _{1}= 1 KT _{2}= 2 KNote how the pressure tripled (from 1 to 3) and the temperature doubled (from 1 to 2).

2) Use the combined gas law to solve:

V_{2}= (P_{1}V_{1}T_{2}) / (P_{2}T_{1})Notice I isolated V

_{2}V

_{2}= [(1) (1) (2)] / [(3) (1)]V

_{2}= 0.67 LIn other words, V

_{2}is two-thirds of V_{1}.

**Solution to (b):**

1) I'll use the form of the combined gas law that isolates V_{2}:

V_{2}= (P_{1}V_{1}T_{2}) / (P_{2}T_{1})

2) The absolute temperature is doubled:

V_{2}= (P_{1}V_{1}(2) / (P_{2}(1)

3) The pressure is reduced by half:

V_{2}= (2) V_{1}(2) / (1) (1)Notice how I went from 2 to 1. I felt that was clearer than going from 1 to 0.5.

4) The result:

V_{2}= 4V_{1}The volume is increased by a factor of 4.

**Solution to (c):**

V_{2}= (P_{1}V_{1}T_{2}) / (P_{2}T_{1})V

_{2}= (1) V_{2}(2) / (2) (1)Note placement of values for temperature, with T

_{1}in the denominator and T_{2}in the numerator. With the symbolic equation being all on one line, you might be tempted to think the temperature was cut in half. Not so!V

_{2}= V_{1}

**Solution to (d):**

1) Place "fake" values into a solution matrix:

P _{1}= 1 kPaP _{2}= 2 kPaV _{1}= 1 LV _{2}= ???T _{1}= 1 KT _{2}= 4 K

2) Use the combined gas law to solve:

V_{2}= (P_{1}V_{1}T_{2}) / (P_{2}T_{1})V

_{2}= (1) V_{1}(4) / (2) (1)V

_{2}= 2V_{1}In other words, the volume doubles.

**Problem #14:** What is molar volume at SATP?

**Solution:**

1) SATP stands for Standard Ambient Temperature and Pressure. It has the following values:

25.0 °C and 100.0 kPaYou may find more discussion here.

2) Let us use the Combined Gas Law to solve this problem. First, a solution matrix:

P _{1}= 101.325 kPaP _{2}= 100.0 kPaV _{1}= 22.414 LV _{2}= ???T _{1}= 273 KT _{2}= 298 KNote the use of values for STP and molar volume at STP.

2) Write the Combined Gas Law, insert values and solve:

P _{1}V_{1}P _{2}V_{2}––––– = ––––– T _{1}T _{2}

(101.325 kPa) (22.414 L) (100.0 kPa) (V _{2})––––––––––––––––––––– = –––––––––––––– 273 K 298 K V

_{2}= [(101.325 kPa) (22.414 L) (298 K)] / [(100.0 kPa) (273 K)]V

_{2}= 24.8 L (to three sig figs)

2) Molar volume at STAP can also be calculated using the Ideal Gas Law:

PV = nRT(100.0 kPa / 101.325 kPa/atm) (V) = (1.00 mol) (0.08206 L atm / mol K) (298 K)

V = 24.8 L

**Problem #15:** A sample of neon has a volume of 0.730 dm^{3} at a temperature of 21.0 °C. and pressure of 102.5 kPa. If the density of neon is 0.900 g/dm^{3} at 0 °C and 101.3 kPa, what is the mass of the sample?

**Solution:**

1) Convert gas conditions to STP:

Here's the cross-multiplied form of the combined gas law:P

_{1}V_{1}T_{2}= P_{2}V_{2}T_{1}(102.5 kPa) (0.730 dm

^{3}) (273 K) = (101.3 kPa) (V_{2}) (294 K)V

_{2}= 0.685887 dm^{3}

2) Determine mass:

0.685887 dm^{3}times 0.900 g/dm^{3}= 0.617 g

**Problem #16:** Suppose the pressure on a 10.0 m^{3} sample gas at 12.0 °C is cut in half.

(a) Is it possible to change the temperature of the gas at the same time such that the volume of the gas doesn't change?

(b) If yes, calculate the new temperature of the gas

**Solution to (a):**

We can't answer this without doing the calculation for (b).This is because the temperature must go down to keep the volume at 10.0 m

^{3}. If the temperature most go down to absolute zero (or below), then the answer would to (a) would be no. Otherwise, we would answer yes.I know the temp must go down because of this: the pressure went down, therefore the volume went up (assuming constant temperature). In order to get the volume back to 10, we must cool the gas down (assuming constant pressure).

**Solution to (b):**

1) Use the combined gas law:

P _{1}V_{1}P _{2}V_{2}––––– = ––––– T _{1}T _{2}

(2 atm) (10.0 m ^{3})(1 atm) (10.0 m ^{3})––––––––––––– = ––––––––––––– 285 K x I used arbitrary values for the pressure. The actual values don't matter, just as long as the pressure gets cut in half.

2) See how the volume does not change? That means it drops out and we have a pressure-temperature relationship:

2 atm 1 atm ––––– = ––––– 285 K x x = 142.5 K

3) Change to Celsius to get −130 °C. So, the answer to (a) is yes.