# Combined Gas Equation Chemistry Tutorial

## Key Concepts

• For a fixed amount of gas, Boyle's Law and Charles' Law can be combined into one equation which is known as the combined gas equation:

 PVT = k (a constant) OR P1V1T1 = P2V2T2 where P1 = initial pressure V1 = initial volume T1 = initial temperature (K) and P2 = final pressure V2 = final volume T2 = final temperature (K)

• Note that:

(i) Pressure can be in any units BUT the units for P1 and P2 MUST be the same.
For example, both are in kPa or both in atm.

(ii) Volume can be in any units BUT the units for V1 and V2 MUST be the same.
For example, both in mL or both in L

(iii) Temperature MUST be in Kelvin (NOT celsius, NOT fahrenheit)

• A gas that obeys the combined gas equation (that is one that obeys Boyle's Law and Charles' Law) is known as an Ideal Gas.

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## Deriving the Combined Gas Equation

Boyle's Law states that, at constant temperature, the volume (V) of a given amount of gas is inversely proportional to its pressure (P):

 V ∝ 1 P

Charles' Law states that, at constant pressure, the volume (V) of the same quantity of gas is directly proportional to its temperature (T) in Kelvin:

V ∝ T (in Kelvin)

So, combining both Boyle's Law and Charles' Law we would say that the volume (V) of a given quantity of gas is directly proportional to its temperature (T) in kelvin and also inversely proportional to its pressure (P):

 V ∝ 1 × T(K)P

that is:

 V ∝ T(K)P

We could use a constant of proportionality (k) to write an equation for this expression:

 V = k × T(K)P

and we could rearrange this equation by multipling both sides by P:

 P × V = P × k × T(K)P P × V = k × T(K)

then divide both sides of the equation by T(K):

 P × V T(K) = k × T(K)T(K) P × V T(K) = k

If we know the pressure (P), the volume (V) and the temperature in Kelvin (T) for a given amount of gas we can find the value of the constant (k).

For example, let's say that we have some cold gas in a syringe.
Initially this gas has a temperature of 250 K and a volume of 20 mL at a pressure of 100 kPa.
We can calculate the value of the constant (k) for this quantity of gas:

 P1 × V1 T1(K) = k

because we know:
P1 = 100 kPa
V1 = 20 mL
T1 = 250 K

so

 100 × 20 250 = k 8 = k

If we now heat the same gas in the syringe to 400 K and allow the gas in the syringe to expand, pushing the plunger (piston) up until the gas occupies a volume of 25 mL, we can calculate the pressure of the gas in the syringe because we have already calculated the value of k for this quantity of gas:

 P2(kPa) × V2(mL) T2(K) = 8

and now
V2 = 25 mL
T2 = 400 K

 P2(kPa) × 25 400 = 8 P2(kPa) × 0.0625 = 8 P2(kPa) × 0.06250.0625 = 8     0.0625 P2(kPa) = 128 kPa

So, under the new conditions of temperature and volume, our gas trapped in the syringe now exerts a pressure of 128 kPa.

Now, we could take a short-cut. Because the value of the constant (k) is the same during the whole of this experiment (because we are using a fixed amount of gas), we could have written:

 P1(kPa) × V1(mL) T1(K) = 8 = P2(kPa) × V2(mL) T2(K)

In other words:

 P1(kPa) × V1(mL) T1(K) = P2(kPa) × V2(mL) T2(K)

The Combined Gas Equation requires that the amount of gas remains constant, and is usually used in this general form:

 P1 × V1 T1(K) = P2 × V2 T2(K)

In which

 P1 = initial pressure of gas P2 = final pressure of gas       (same units as P1) V1 = initial volume of gas V2 = final volume of gas       (same units as V1) T1 = initial temperature of gas       (in Kelvin) T2 = final temperature of gas       (in Kelvin)

We need to know 5 of the 6 values in order to calculate the 6th value:

To calculate initial pressure (P1):

 P1 = P2 × V2 × T1 V1 × T2

To calculate final pressure (P2):

 P2 = P1 × V1 × T2 V2 × T1

To calculate initial volume (V1):

 V1 = P2 × V2 × T1 P1 × T2

To calculate final volume (V2):

 V2 = P1 × V1 × T2 P2 × T1

To calculate initial temperature (T1):

 T1 = P1 × V1 × T2 P2 × V2

To calculate final temperature (T2):

 T2 = P2 × V2 × T1 P1 × V1

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## Worked Example: Combined Gas Equation to Calculate Gas Pressure

Question: A quantity of gas occupied a volume of 1.00 L at 1.00 atm pressure and 80.0oC.
What pressure is required to compress the gas to a volume of 500 mL at 40.0oC?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate final pressure
P2 = ? atm

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions: Assume a constant amount of gas (that is, no gas was lost and no gas was added during the experiment)

P1 = 1.00 atm

V1 = 1.00 L

T1 = 80.0°C
Convert T in °C to K (temperature MUST be in Kelvin)
T1(K) = 80.0°C + 273 = 353 K

V2 = 500 mL
Convert volume in mL to L so that both volumes are in the same units (L)
V2 = 500 mL ÷ 1000 mL/L = 0.500 L

T2 = 40.0°C
Convert T in °C to K (temperature MUST be in Kelvin)
T2(K) = 40.0°C + 273 = 313 K

3. What is the relationship between what you know and what you need to find out?
Because the experiment is done with a constant quantity of gas we can use the Combined Gas Equation:

 P1V1T1 = P2V2T2

Which we can rearrange to find P2:

 P2 = P1V1T2V2T1

4. Substitute the values into the equation and solve for P2:

 P2 = P1V1T2V2T1 = 1.00 × 1.00 × 3130.500 × 353 = 313 176.5 = 1.77 atm

Consider, the volume of the gas has halved (compressed from 1000 mL to 500 mL), so, if the temperature was constant we would expect the pressure to double, that is 1 atm would increase to 2 atm. (Using Boyle's Law which states that for a given amount of gas at constant temperature the volume is inversely proportional to pressure)

But, the temperature was not constant, it fell from 353K to 313 K, so we expect the pressure of the gas to be less than 2 atm (slower moving gas particles will exert less pressure, see Kinetic Theory of Gases).
The final pressure is likely to be 313/353 × 2 atm = 1.77 atm
Since this value agrees with the value we calculated above, we are reasonably confident that our answer is correct.

6. State your solution to the problem "final gas pressure":

P2 = 1.77 atm

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## Worked Example: Combined Gas Equation to Calculate Gas Volume

Question: A quantity of gas has a volume of 24.5 L at 101.3 kPa and 298 K.
What volume will this gas occupy if it is cooled to 0.00oC at a pressure of 2.00 atm?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate final volume of gas
V2 = ? L

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions: assume constant amount of gas (no gas is lost nor added to the systen during the experiment)

P1 = 101.3 kPa

V1 = 24.5 L

T1 = 298 K

P2 = 2.00 atm
Convert pressure in atm to kPa so both pressures are in the same units
1 atm = 101.3 kPa
2 atm = 2 × 101.3 = 202.6 kPa
P2 = 202.6 kPa

T2 = 0.00°C
Convert T in °C to K (temperature MUST be in Kelvin)
T2(K) = 0.00°C + 273 = 273 K

3. What is the relationship between what you know and what you need to find out?
Because the experiment is done with a constant quantity of gas we can use the Combined Gas Equation:

 P1V1T1 = P2V2T2

Which we can rearrange to find V2:

 V2 = P1V1T2P2T1

4. Substitute the values into the equation and solve for V2

 V2 = P1V1T2P2T1 = 101.3 × 24.5 × 273202.6 × 298 = 677545.0560374.8 = 11.2 L

Consider Charles' Law, for a given amount of gas at constant pressure, the volume of the gas is directly proportional to its temperature.
Temperature of our gas decreases from 298 K to 273 K, so we expect the volume to fall
New volume = 273/298 × 24.5 = 22.4 L

But the pressure was not constant, the pressure doubled from 101.3 kPa (1 atm) to 2 atm, so we predict that the final volume will be half that we expect (because Boyle's Law tells us the volume is inversely proportional to pressure), that is,
V = ½ × 22.4 = 11.2 L

Since this value for volume agrees with our calculation above, we are reasonably confident that our answer is plausible.

6. State your solution to the problem "final gas volume":

V2 = 11.2 L

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## Worked Example: Combined Gas Equation to Calculate Gas Temperature

Question: A quantity of gas has volume of 2.50 L at 760 mm Hg and 20.0oC.
If the gas is compressed to a volume of 750 mL by a pressure of 2.00 atm what is its temperature in Kelvin?

Solution:

(Based on the StoPGoPS approach to problem solving.)

1. What is the question asking you to do?

Calculate final temperature in Kelvin
T2 = ? K

2. What data (information) have you been given in the question?

Extract the data from the question:

Conditions: assume constant amount of gas (gas is neither lost nor added to the system during the experiment)

P1 = 760 mm Hg

V1 = 2.50 L

T1 = 20.0°C
Convert T in °C to K (temperature MUST be in Kelvin)
T1(K) = 20.0°C + 273 = 293 K

P2 = 2.00 atm
Convert pressure in atm to mm Hg (so both pressure units are the same)
1 atm = 760 mm Hg
2 atm = 2.00 × 760 = 1520 mm Hg
P2 = 1520 mm Hg

V2 = 750 mL
Convert volume in mL to L so that both volumes are in the same units (L)
V2 = 750 mL ÷ 1000 mL/L = 0.750 L

3. What is the relationship between what you know and what you need to find out?
Because the experiment is done with a constant quantity of gas we can use the Combined Gas Equation:

 P1V1T1 = P2V2T2

Which we can rearrange to find T2:

 T2 = P2V2T1P1V1

4. Substitute the values into the equation and solve for T2

 T2 = 1520 ×0.750 × 293760 × 2.50 = 3340201900 = 176 K

Consider Boyle's Law which states that for a given amount of gas at constant temperature, PV = constant
A proportion of this constant is made up of temperature, so "constant" = T × k
initially: P1V1 = 760 × 2.50 = 1900 = T1k
finally: P2V2 = 1520 × 0.75 = 1140 = T2k

Therefore we could write:
k = 1900 ÷ T1
k = 1140 ÷ T2

So,
1900 ÷ T1 = 1140 ÷ T2
1900 ÷ 293 = 1140 ÷ T2
6.48 = 1140 ÷ T2
T2 = 1140 ÷ 6.48 = 176 K

Since this value agrees with the value for final temperature we calculated above, we are reasonably confident that our answer is plausible.

6. State your solution to the problem "final temperature in Kelvin":

T2 = 176 K

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