The Combined Gas Law

Manipulating Variables in equations Often in an equation we want to isolate some variable, usually the unknown From math: what ever you do to one side of an equation you have to do to the other side Doing this keeps both sides the same E.g. x + 5 = 7, what does x equal? We subtract 5 from both sides … x + 5 – 5 = 7 – 5, thus x = 2 Alternatively, we can represent this as 5 moving to the other side of the equals sign … x + 5 = 7 becomes x = 7 – 5 or x = 2 Thus, for addition or subtraction, when you change sides you change signs

Multiplication and division We can do a similar operation with multiplication and division E.g. 5x = 7, what does x equal? We divide each side by 5 (to isolate x) … 5x/5 = 7/5 … x = 7/5 … x = 1.4 Alternatively, we can represent this as 5 moving to the other side of the equals sign … 5x = 7 becomes x = 7/5 Thus, for multiplication and division, when you change sides you change position (top to bottom, bottom to top)

Multiplication and division Let’s look at a more complicated example: (x) (y) 5 = 7a b Isolate a in the equation: Move b to the other side (from bottom to top) 5 b (x) (y) = 7a Move 7 to the other side (from top to bottom) (x)(y)(b) 5 = 7a (x)(y)(b) (35) = a (x)(y)(b) (5)(7) = a or

Multiplication and division This time, isolate b in the equation: (x) (y) 5 = 7a b Move b to the other side (it must be on top) … (x) (y) 5 = 7a b Move everything to the other side of b 35a xy = b (b)(x)(y) 5 = 7a Q - Rearrange the following equation to isolate each variable (you should have 6 equations) P1V1 P2V2 T1 T2 =

Combined Gas Law Equations P1 = P2T1V2 T2V1 P2 = P1T2V1 T1V2 T1 = P1T2V1 P2V2 T2 = P2T1V2 P1V1 V1 = P2T1V2 T2P1 V2 = P1T2V1 P2T1

These are all subsets of a more encompassing law: the combined gas law Combining the gas laws So far we have seen two gas laws: Robert Boyle Jacques Charles Joseph Louis Gay-Lussac V1 T1 = V2 T2 P1 T1 = P2 T2 P1V1 = P2V2 These are all subsets of a more encompassing law: the combined gas law P1V1 P2V2 T1 T2 =

Q1 What is the pressure of 12.5 mL of gas at room temperature if the same gas occupies 50.0mL at atmospheric pressure? (101kPa)

Q 1 V1 = 50.0 ml, P1 = 101 kPa V2 = 12.5 mL, P2 = ? T1 = T2 P1V1 T1 = (101 kPa)(50.0 mL) (T1) = (P2)(12.5 mL) (T2) (101 kPa)(50.0 mL)(T2) (T1)(12.5 mL) = 404 kPa = (P2) Notice that T cancels out if T1 = T2

Q2 If a 0.10 L balloon at room temperature (25oC) is heated to 190oC, what will be its new volume?

Q 2 V1 = 0.10 L, T1 = 298 K V2 = ?, T2 = 463 P1 = P2 P1V1 T1 = P2V2 T2 (P1)(0.10 L) (298 K) = (P2)(V2) (463) (P1)(0.10 L)(463 K) (P2)(298 K) = 0.16 L = (V2) Notice that P cancels out if P1 = P2

Q3 If a cylinder contains a gas at 150kPa at 35oC, what must be the temperature if the pressure changes to 250kPa?

Q 3 P1 = 150 kPa, T1 = 308 K P2 = 250 kPa, T2 = ? V1 = V2 P1V1 T1 = (150 kPa)(V1) (308 K) = (250 kPa)(V2) (T2) (250 kPa)(V2)(308 K) (150 kPa)(V1) = 513 K = 240 °C = (T2) Notice that V cancels out if V1 = V2

Q4 If a gas is found in a 5.0L container at room temperature and 100kPa, what will be the volume of the gas if the pressure drops to 90kPa and the temperature rises to 35oC?

Q 4 P1 = 100 kPa, V1 = 5.00 L, T1 = 293 K P2 = 90 kPa, V2 = ?, T2 = 308 K P1V1 T1 = P2V2 T2 (100 kPa)(5.00 L) (293 K) = (90 kPa)(V2) (308 K) (100 kPa)(5.00 L)(308 K) (90 kPa)(293 K) = 5.84 L = (V2) Note: although kPa is used here, any unit for pressure will work, provided the same units are used throughout. The only unit that MUST be used is K for temperature.

Q5 A gas is found to occupy 1.0L of space at 30oC and 800kPa. What will be the volume of the gas if the temperature is decreased to room temperature and the pressure is decreased to 100kPa?

Q 5 P1 = 800 kPa, V1 = 1.0 L, T1 = 303 K P2 = 100 kPa, V2 = ?, T2 = 298 K P1V1 T1 = P2V2 T2 (800 kPa)(1.0 L) (303 K) = (100 kPa)(V2) (298 K) (800 kPa)(1.0 L)(298 K) (100 kPa)(303 K) = 7.9 L = (V2)

Q6 A gas at 6.5 atm and 10oC occupies 2.0 mL of space. It is then placed in a lower pressure of 0.95 atm and heated to 24oC. What will be its new volume?

Q 6 P1 = 6.5 atm, V1 = 2.0 mL, T1 = 283 K P2 = 0.95 atm, V2 = ?, T2 = 297 K P1V1 T1 = P2V2 T2 (6.5 atm)(2.0 mL) (283 K) = (0.95 atm)(V2) (297 K) (6.5 atm)(2.0 mL)(297 K) (0.95 atm)(283 K) = 14 mL = (V2)

Q7 How many moles of gas will there be relative to the original situation? Ans: same amount. Equations only work if gas is not allowed to leave the container.