James Caudle
Wendy Cook
John Putnam


Connects how Temperature relates to Volume of gases. (Pressure is held constant)

This is a direct proportion (one goes up the other goes up)
V
1
T
1

V
2
T
2


Connects how Pressure relates to Volume of gases. (Temperature is held constant)

This is an inverse proportion (one goes up the other goes down)
P
1
V
1

P
2
V
2


Connects how Pressure relates to
Temperature (Kelvin scale) of gases. (Volume is held constant)

This is a direct proportion (one goes up the other goes up)
P
1
T
1

P
T
2
2


Take a piece of paper and fold in in half length wise.

With it folded still, do it again.

Now evenly space out P, T, and V in that order (very important that it is in alphabetical order)


Pinch your finger over the P (Pivot point)

Now raise the other end. This tells you that as T increases so does V

Pinch your finger on T (Pivot point)

Now raise the P on the end. This tells you that as P increases then V decreases and vice versa

Pinch your finger on V (Pivot point)

Now raise the other end. This tells you that as P increases so does T


A balloon filled with helium gas has a volume of 500 mL at a pressure of 1 atm. The balloon is released and reaches and altitude of 6.5 km, where the pressure is 0.5 atm. If the temperature has remained the same, what does the gas occupy at this height?

#1 Which law would you use?

#2 How do you know?

#3 Anything need to be converted?

#4 Solve (don’t forget sig figs & labels)


#1 Boyle’s Law

# 2 Specifically says “Temp has remained the same” (or your PTV ruler)

#3 No


#4 P
1
V
1

P
2
V
2
( 1 atm )( 500 mL )

( 0 .
5 atm )( V
2
)
( 500 atm * mL )

V
2
( 0 .
5 atm )
1000 mL

V
2


A sample of helium gas has a pressure of 1.20 atm at 22 °C. At what Celsius temperature will the helium reach a pressure of 2.00 atm, assuming constant volume?

#1 Which law would you use?

#2 How do you know?

#3 Anything need to be converted?

#4 Solve (don’t forget sig figs & labels)


#1 Amontons

#2 Says “assuming constant volume” (or your
PTV ruler)

#3 Yes, Temp needs to be in Kelvin. (add
273.15)


#4
P
1
T
1

P
T
2
2
1 .
20 atm

295 K
590 .
atm * K
2 .
00 atm
T
2

1 .
20 atmT
2
590 .
atm * K

1 .
20 atm
492 K

492

T
2
273

219

C


A sample of oxygen that occupies 1.00 x 10 6 mL at 575 mmHg is subjected to a pressure of 1.25 atm. What will the final volume be if the temperature is held constant?

#1 Which law would you use?

#2 How do you know?

#3 Anything need to be converted?

#4 Solve (don’t forget sig figs & labels)


#1 Boyle’s Law

# 2 Specifically says “Temp has remained the same” (or your PTV ruler)

#3 Yes, mmHg needs turned into atm or vice versa—Don’t forget dimensional analysis
575 mmHg
1
1 atm x
760 mmHg

575
760 atm or
1 .
25 atm
1 x
760 mmHg
1 atm

1 .
25 x 760 mmHg


#4
P V
1 1

P
2
V
2
( 575 mmHg )( 1 .
00 x 10
6 mL )

(( 1 .
25 x 760 ) mmHg ) V
2
6 .
05 x 10
5 mL or
P
1
V
1

P
2
V
2
(( 575 / 760 ) atm )( 1 .
00 x 10
6 mL )

( 1 .
25 atm ) V
2
6 .
05 x 10
5 mL


A sample of nitrogen gas is contained in a piston with a freely moving cylinder. At
0.0°C, the volume of the gas is 375 mL. To what temperature must the gas be heated to occupy a volume of 500.0 mL?

#1 Which law would you use?

#2 How do you know?

#3 Anything need to be converted?

#4 Solve (don’t forget sig figs & labels)


#1 Charles’ Law

#2 Pressure is not mentioned (or PTV ruler)

#3 Yes, Temp needs to be in Kelvin. (add
273.15)


#4
V
1
T
1

V
2
T
2
375 mL
273 .
2 K

500 .
0 mL
T
2
( 500 .
0 mL )( 273 .
2 K )

( 375 mL ) T
2
( 500
364 K
.
0 mL )( 273 .
2 K )

T
2
( 375 mL )

364

273

91

C


If you noticed each of the three laws use only two of three variables. Because of this we can combine all three laws into one.
P
1
V
1
T
1

P
2
V
2
T
2

We can do any of the four examples we just did using this by simply removing the variable that remains constant.


Nature does not keep one variable constant most of the time. As one property changes it can change not just one other property but both. Especially if the container is an elastic container that can change its shape.


A 700.0 mL gas sample at STP is compressed to a volume of 200.0 mL, and the temperature is increased 30.0 ºC. What is the new pressure of the gas in Pa?

#1 What is STP?

#2 Anything need to be coverted?

#3 Solve (don’t forget sig figs & labels)


#1 Standard temp is 273.15 K and standard pressure is 101325 Pa

#2 Yes, Temp needs to be in Kelvin. (add
273.15)


#3
( 101325 Pa )(
( 101325 Pa )(
700
273 .
2 K
.
0 mL )

P
700 .
0 mL )( 303 .
2
(
2
200
K )
.
0
303 .
2
 mL )
K
P
2
( 200 .
0 mL )( 273 .
2 K )
( 101325 Pa )( 700 .
0 mL )( 303 .
2 K )

P
2
P
2
( 200 .
0 mL )( 273 .
2 K )

39360 Pa