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Unit 9: Gases Ideal Gas Law. After today you will be able to… Explain what an ideal gas is Calculate an unknown pressure, temperature, volume, or amount.

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Presentation on theme: "Unit 9: Gases Ideal Gas Law. After today you will be able to… Explain what an ideal gas is Calculate an unknown pressure, temperature, volume, or amount."— Presentation transcript:

1 Unit 9: Gases Ideal Gas Law

2 After today you will be able to… Explain what an ideal gas is Calculate an unknown pressure, temperature, volume, or amount of gas using the ideal gas law equation

3 Ideal Gases gases that are said to follow all assumptions of kinetic molecular theory. Ideal gases are gases that are said to follow all assumptions of kinetic molecular theory. conform to all gas laws. An ideal gas is also considered to conform to all gas laws.

4 Ideal Gases Up until now we have only changed variables such as pressure, volume, and temperature of a gas. amount of gas was assumed constant. In each of these cases the amount of gas was assumed constant. amount of gas by including the variable, n. The combined gas law can be modified to include the amount of gas by including the variable, n.

5 Ideal Gas Law This gas law relates the amount of gas (in moles) to the volume it would occupy at a particular temperature and pressure.

6 Ideal Gas Law Where, pressure (atm) P= pressure (atm) volume (L) V= volume (L) moles (mol) n= moles (mol) 0.0821 L · atm/mol · K R= 0.0821 L · atm/mol · K temperature (K) T= temperature (K) PV=nRT Ideal Gas Constant R is called the Ideal Gas Constant (it has multiple values, but for our purposes we will only use this one). STOP! It is often called the “picky” law! The units must be what you see here!

7 Ideal Gas Law Example At what pressure would 0.212 mol of a gas occupy 6.84L at 89°C? P= V= n= R= T= ? 6.84L 0.212mol 0.0821L · atm/mol · K (P)(6.84L)= (0.212mol) P = 0.92atm PV=nRT (0.0821) 89°C + 273= 362K (362K)

8 Ideal Gas Law Example At what temperature would 52.3g of methane (CH 4 ) gas occupy 65.7L at 184kPa? P= V= n= R= T= 184kPa 65.7L 52.3gCH 4 0.0821 L · atm/mol · K (1.82atm)(65.7L)= (3.26mol) T = 447K PV=nRT (0.0821) ? (T) x 101.3kPa 1 atm =1.82 atm x 16.05gCH 4 1 mol CH 4 = 3.26 mol CH 4


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