2 Manipulating Variables in equations Often in an equation we want to isolate some variable, usually the unknownFrom math: what ever you do to one side of an equation you have to do to the other sideDoing this keeps both sides the sameE.g. x + 5 = 7, what does x equal?We subtract 5 from both sides …x + 5 – 5 = 7 – 5, thus x = 2Alternatively, we can represent this as 5 moving to the other side of the equals sign …x + 5 = 7 becomes x = 7 – 5 or x = 2Thus, for addition or subtraction, when you change sides you change signs
3 Multiplication and division We can do a similar operation with multiplication and divisionE.g. 5x = 7, what does x equal?We divide each side by 5 (to isolate x) …5x/5 = 7/5 … x = 7/5 … x = 1.4Alternatively, we can represent this as 5 moving to the other side of the equals sign …5x = 7 becomes x = 7/5Thus, for multiplication and division, when you change sides you change position (top to bottom, bottom to top)
4 Multiplication and division Let’s look at a more complicated example:(x) (y)5=7abIsolate a in the equation:Move b to the other side (from bottom to top)5b(x) (y)=7aMove 7 to the other side (from top to bottom)(x)(y)(b)5=7a(x)(y)(b)(35)=a(x)(y)(b)(5)(7)=aor
5 Multiplication and division This time, isolate b in the equation:(x) (y)5=7abMove b to the other side (it must be on top) …(x) (y)5=7abMove everything to the other side of b35axy=b(b)(x)(y)5=7aQ - Rearrange the following equation to isolate each variable (you should have 6 equations)P1V1 P2V2T T2=
6 Combined Gas Law Equations P1=P2T1V2T2V1P2=P1T2V1T1V2T1=P1T2V1P2V2T2=P2T1V2P1V1V1=P2T1V2T2P1V2=P1T2V1P2T1
7 These are all subsets of a more encompassing law: the combined gas law Combining the gas lawsSo far we have seen two gas laws:Robert BoyleJacques CharlesJoseph Louis Gay-LussacV1T1=V2T2P1T1=P2T2P1V1=P2V2These are all subsets of a more encompassing law: the combined gas lawP1V1 P2V2T T2=Read pages 437, Do Q 26 – 33 (skip 31)
8 Q 26 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
11 Q 27 V1 = 0.10 L, T1 = 298 K V2 = ?, T2 = 463 P1 = P2 P1V1 T1 = P2V2 (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
12 Q 28 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
13 Q 29P1 = 100 kPa, V1 = 5.00 L, T1 = 293 KP2 = 90 kPa, V2 = ?, T2 = 308 KP1V1T1=P2V2T2(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.
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