1.
Which law compares gases at two different sets of conditions, with temperature as a constant and pressure and volume as variables?
Correct Answer
B. Boyle's Law
Explanation
Boyle's law: P1V1 = P2V2 Temperature is constant.
(Where 1 indicates the first set of conditions and a 2 indicates the second set of conditions.
{Charle's law is V1/T1 = V2/T2 with pressure constant;
Ideal is PV = nRT and
combined is PV/T = PV/T}
2.
If pressure of a gas is 1.2 atmospheres with a volume of 12 L, how much pressure is exerted by the gas if the volume is changed to 9L? (Assume constant temperature)
Correct Answer
D. 1.6 L
Explanation
If the variables are only Pressure and Volume, (with constant temperature), this indicates Boyle's Law:
P1V1 = P2 V2 (Where 1 indicates the first set of conditions and 2 indicates the second set of conditions).
(1.2 atm) (12L) = (P2) (9L)
Math: 1.2 times 12 divided by 9.
3.
If the pressure of a gas is 103.2 kPa at a temperature of 100 Celsius and a volume of 4L is changed to a lower pressure of 80 kPa and 100 degrees celsius, what will be the new volume? (Hint: kPa is kilo pascals -- a unit of measure for pressure).
Correct Answer
C. 5.16 L
Explanation
Since the temperature is a constant (it started at 100C and ended at 100C) and Pressure and volume are the variables, this is Boyle's Law. P1V1 = P2V2. (Where 1 indicates the initial set of conditions and 2 indicates the final set of conditions).
(103.2kPa) (4L) = (80 kPa) (V2)
Math: 103.2 x 4 divided by 80
4.
A gas at a temperature of 80Celsius and a volume of 3.2 Liters is brought into an area with the temperature of 100 Celsius, what is the new volume if the pressure remains constant?
Hint 1: This is Charles law (V1/T1 = V2/T2)
Hint 2: ALL GAS PROBLEMS MUST use the KELVIN temperature scale.
(To change from Celsius to Kelvin: add 273)
Correct Answer
A. 3.38 L
Explanation
(3L)/(353K) = V2/(373K)
Math: 3 x 373 divided by 353.
5.
A gas at a temperature of 34 Celsius and a volume of 6.3 Liters is brought into a new area where the gas expands to a volume of 8.0 Liters. What is the new temperature? (Assume constant Pressure)
Correct Answer
B. 389.84 K
Explanation
This problem uses Charles' law. (V1/T1 = V2/T2)
MAKE SURE YOU CHANGE YOUR TEMPERATURE TO KELVIN (add 273)!!
(6.3L)/307K = (8L)/T2
Math: 8 x 307 divided by 6.3
6.
A gas in a balloon measures 2.2 L 1.1 atmospheres and 32 degrees Celsius. The environment is changed and the balloon is now 1.8 L and 1.4 atmospheres. What is the new temperature?
Hint 1: Always change temperature to Kelvin scale.
Hint 2:
Pressure volume and temperature are all changed.
Use the combined gas law:
Correct Answer
A. 317.6 K
Explanation
The combined gas law states that the product of pressure and volume is directly proportional to the temperature. In this case, we have the initial pressure of 1.1 atmospheres and volume of 2.2 L at a temperature of 32 degrees Celsius. The new pressure is 1.4 atmospheres and volume is 1.8 L. To find the new temperature, we can set up the equation (P1 * V1) / T1 = (P2 * V2) / T2, where P1, V1, and T1 are the initial values and P2, V2, and T2 are the new values. Rearranging the equation, we get T2 = (P2 * V2 * T1) / (P1 * V1). Plugging in the values, we get T2 = (1.4 * 1.8 * 305.15) / (1.1 * 2.2) = 317.6 K.
7.
A gas in a balloon measures 3.6 L, 0.9 atmospheres and 34 degrees Celsius. The environment is changed and the balloon is now 1.4 atmospheres at a temperature of 38 degrees Celsius. What is the new volume?
Correct Answer
D. 2.34 L
Explanation
The ideal gas law states that the product of pressure and volume is directly proportional to the product of temperature and the number of moles of gas. In this case, the initial pressure, volume, and temperature are given as 0.9 atm, 3.6 L, and 34 degrees Celsius respectively. The final pressure and temperature are given as 1.4 atm and 38 degrees Celsius. To find the new volume, we can use the formula P1V1/T1 = P2V2/T2, where P1, V1, and T1 are the initial pressure, volume, and temperature respectively, and P2, V2, and T2 are the final pressure, volume, and temperature respectively. Plugging in the given values, we can solve for V2, which comes out to be approximately 2.34 L.
8.
What is the volume of 2 moles of Helium at a temperature of 32 degreees celsius and a pressure of 1.1 atmospheres ?
Given R = 0.08321 Latm/molK and 8.314 kPa L/mol K
Hint: There is only 1 set of conditions for this gas problem and the moles of the gas are given.
This indicates the ideal gas law: PV = nRT.
Remember that n is equal to the number of moles.
P is the pressure, V is the volume and T is the temperature (change to Kelvin).
The R MUST match the units in the rest of the problem.
IF pressure is given in atmosphere, then use R = 0.08321 Latm/molK .
IF pressure was given in kPa, then use 8.314 Kpa L/mol K
Note also that volume MUST be in Liters and Temperature (as for all gas problems) must be in Kelvin.
Correct Answer
C. 46.14 L
Explanation
(1.1 atm) (V) = (2 moles) (0.08321 L atm/molK) (305K)
9.
What is the temerature of 3.4 moles of Helium at a pressure of 1.1 atmospheres and a volume of 3.7 Liters. (What is the temperature in Kelvin?)
Given R = 0.08321 Latm/molK and 8.314 kPa L/mol K
Correct Answer
A. 14.39 K
Explanation
(1.1 atm) (3.7 L) = (3.4 moles) (0.08321 Latm/mol K) (T)
10.
How many moles is present in a container that is 79.9 Liters with a pressure of 105 kPa and a temperature of 24 degrees Celsius?
Given R = 0.08321 Latm/molK and 8.314 kPa L/mol K
Correct Answer
C. 3.4 moles
Explanation
There is only 1 set of conditions and you are solving for moles, therefore use the ideal gas law:
PV = nRT.
Since Pressure is given in kPa, you must use R = 8.314 kPa L/mol K.
Make sure to change to Kelvin for temperature.
(105 kPa) (79.9L) = n (8.314 kPa L/mol K) (297K)