Agas occupies a 1.5 l container at 25 degrees celsius and 2.0 atm. if the gas is transferred to a 3.0 l container at the same temperature, what will be its new pressure?
a) 3.0 atm
b) 2.0 atm
c) 1.0 atm
d) 4.0 atm

To solve this we assume that the gas is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant temperature and number of moles of the gas the product of PV is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

P1V1 =P2V2

P2 = P1 x V1 / V2

P2 = 2.0 x 1.5 / 3

P2 = 1 atm

Option c. 1 atm

Explanation:

We can do this exercise in two ways.

1. using the ideal gas equation which is PV = nRT

2. Using boyle's law.

Now, with the first option, would be longer, because we will need to calculate the number of moles present in the first container, and then, replace it in the second container to calculate P.

However, in this case, is easier to use Boyle's law.

Boyle's law is a relation concerning the compression and expansion of a gas at constant temperature.

In this case, we can see that the gas, is being expanded to a major volume, but temperature is constant (same 25 °C), therefore, we can use the following expression to calculate the new pressure:

P1*V1 = P2*V2

Now, let's replace the data provided in this expression, and solve for P2:

P2 = P1*V1 / V2

P2 = 2 * 1.5 / 3

P2 = 3 / 3

P2 = 1 atm

This would be the new pressure of the gas at constant temperature

1. the question asks the most appropriate, which seems to rule out "all of the above." your choice appears to be the correct one.

2. the protocol described is good for identifying correlation, but not cause and effect. for cause/effect, you need to devise a protocol that will assign the treatment (or not) to groups that have essentially the same "demographics" with respect to all the other variables. your choice regarding control group seems appropriate. random assignment of treatment (choice d) may be appropriate also. (i think you can argue that b and d are just different descriptions of the same method.)

3. d will randomize the birds, but may not guarantee there are 10 in each group. e will do both, so seems the better choice.

comment on question 2

based on the problem statement, it seems clear that confident, attentive drivers are more likely to use daytime running lights. so, the protocol might need to be devised in such a way that the lights would be turned on or off without the driver's knowledge. (an attentive driver will figure it out.) assigning a confident, attentive driver to a group required to drive without running lights might mess with his/her confidence and attentiveness. you don't want your experiment to do that. (nor do you want the experiment to subject a driver to greater risk of accident as a consequence of driving without running lights.) experiments involving human safety are always problematic.

what is the average time it took the car to travel 0.25 meters?

⇒ 2.23 seconds

what is the average time it took the car to travel 0.50 meters?

⇒ 3.13 seconds