Answers for the problems
in ppt. #1:
*1. A gas at 45°C and 1.2 atm
is cooled to a temperature of
5°C. What is the new
pressure of the gas?
Set this up and work it
before you check it on the
next slide.
T1 =
P1 =
T2 =
P2 = ?
1.05 atm
45°C + 273 = 318 K
5°C + 273 = 278 K
1.2 atm
318
=
Gay-Lussac’s Law
Direct Relationship
P2
278
*2. A 2.3L sample of gas
under 500 torr of pressure
is allowed to expand with a
final pressure of 2.3 atm.
What is the final volume?
**Notice both torr and atm are used
V1 =
P1 =
V2 = ?
P2 =
.658L
First find the torr in the 2.3 atm:
2.3 atm x 760 torr = 1748 torr
1 atm
500 x 2.3 = 1748 x V2
Boyle’s Law
Inverse Relationship
*3. A gas at a pressure of
700 torr is pressurized
to 1200 torr and 50°C.
What was the initial
temperature of the gas in
degrees Celsius?
You will need to find it in Kelvin first.
T1 = ?
P1 =
T2 =
P2 =
-84.6°C
50°C + 273 = 323
500
T1
=
1200
323
Gay-Lussac’s Law
Direct Relationship
*4.
A syringe containing
5.5 mL of gas at
standard temperature
is allowed to expand to
twice its volume. What
is the final
temperature?
What is twice the 5.5mL?
V1 =
T1 =
V2 =
T2 = ?
546K
0°C + 273 = 273 K
V2 = 2 x 5.5 = 11 mL
5.5
273
=
11
T2
Charles’ Law
Direct Relationship
*5. 15L of a gas at
standard pressure is
shrunk to a volume of
12.5L. What is the new
pressure, in torr?
V1 =
P1 =
V2 =
P2 = ?
912 torr
Standard pressure in torr =
760
760 x 15 = P2 x 12.5
Boyle’s Law
Inverse Relationship
The combined gas law:
If all three of these variables P, V and T
are changed in a new set of conditions
(for instance STP) and 5 of the six
variables are known, the sixth can be
calculated using….THE COMBINED
GAS LAW
P1V1 = P2V2
T1
T2
*6. What is the pressure
in mmHg exerted by a
11.2L sample of nitrogen
gas in a 10.0L container
at 298K?
P1=?
V1= 10.0L
T1= 298K
P2= 760 mmHg
V2= 11.2L
T2= 273K
P1= 760(11.2)298 = 929 mmHg
10.0(273)
*7. What is the pressure in a
15.0L tank containing 2550L
of propane (C3H8) to grill
brats on a warm summer day
of 40.0C (…wouldn’t that be
a small miracle!)?
P1=?
V1 =15.0L
T1 =313K
P2 =760torr
V2= 2550L
T2=272K
P1 =760(2550)313 = 149,000torr
15.0(272)
*8. Calculate the volume of
a gas at STP that
occupies 255mL at 2.2
atm in a lab that is 25.5
degrees C.
P1= 2.2 atm
V1= 255 mL
T1= 25.5 + 273
= 298.5K
V2 = 2.2(255)273 = 513 mL
298.5(1)
P2 = 1 atm
V2 = ?
T2 = 273K