A balloon filled with helium has a volume of 30.0 L at a pressure of 100 kPa and a temperature at 15.0 Celsius. What will The volume of the balloon be if the temperature is increased at 80.0 Celsius and the pressure changes to 115 kPa?

The answer to your question is V2 = 31.97 L

Explanation:

Data

Volume 1 = 30 L

Pressure 1 = 100 kPa

Temperature 1 = 15°C

Volume 2 = ?

Pressure 2 = 115 kPa

Temperature 2 = 80°C

Process

1.- Convert temperature to °K

Temperature 1 = 15 + 273 = 288°K

Temperature 2 = 80 + 273 = 353°K

2.- Use the Combined gas law to solve this problem

                  P1V1/T1 = P2V2/T2

-Solve for V2

                 V2 = P1V1T2 / T1P2

-Substitution

                 V2 = (100 x 30 x 353) / (288 x 115)

- Simplification

                 V2 = 1059000 / 33120

- Result

                 V2 = 31.97 L

The new volume is 31.97 L

Explanation:

Step 1: Data given

The initial volume = 30.0 L

The initial pressure = 100 kPa

Temperature = 15.0 °C = 288 K

The temperature is increased to 80.0 °C = 353 K

The pressure increases to 115 kPa

Step 2: Calculate the new volume

P1*V1 /T1  = P2*V2/T2

⇒with P1 = the initial pressure = 100 kPa

⇒with V1 = the initial volume = 30.0 L

⇒with T1 = the initial temperature = 288 K

⇒with P2 = the increased pressure = 115 kPa

⇒with V2 = the new volume = TO BE DETERMINED

⇒with T2 = the increased temperature = 353 K

(100 * 30.0)/288 = (115 * V2)/353

10.417 = (115 * V2)/353

V2 = 31.97 L

The new volume is 31.97 L

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.

it is the most commonly accepted one