AT&T; 11:02 AM session try.com largest volume ANSWER Figure I Figure 3 Figure 2

Question

AT&T; 11:02 AM session try.com largest volume ANSWER Figure I Figure 3 Figure 2 Watch the video that shows the relationship between pressure and volume for a particular gas. Robert Boyle was the first one to document the change in the pressure of a gas when its volume is increased or decreased at a constant temperature and the change in the volume of a gas when its pressure is increased or decreased at a constant temperature Part A Watch the video to determine which of the following relationships are correct according to Boyle's law Check all that apply. You did nor net open hints for this part ANSWER: Part B Standard temperature and pressure (STP) are considered to be 273 K and 1.0atm Predict which of the following changes will cause the volume of the balloon to increase or decrease assuming that the temperature and the gas filling the balloon remain unchanged. Drag the appropriate items to their respective bins You did not open hints for this part. ANSWER This content is not available on your mobile device, but your work is still incomplete. Please use a dfferent device to submit your solution for grading Part C A certain gas is present in a 10.0 L cylinder at 3.0 atm pressure. If the pressure is increased to 6.0 .de volume of the gas decreases to 50L. Find the two constants&,the initial value of a, and &, the final value of a, to verify whether the gas obeys Boyle's law Express your answers to two significant figures separated by a comma.

Explanation / Answer

Part A : The relation between P and V is,

P = 1/V

Part B. We have the equation,

V = nRT/P

The volume of balloon would increase with decreasing pressure and would increase with increasing temperature.

Part C. For initial k1 = P1V1 = 3 x 10 = 30

For final k2 = P2V2 = 6 x 5 = 30   

So the values does obey Boyle's law PV = constant

Part D. Using,

P1V1 = P2V2

with,

P1 = 7.5 atm

V1 = 15 L

P2 = ?

V2 = 3.8 L

we get,

final pressure P2 = 7.5 x 15/3.8 = 29.60 atm

8.11

Part A : The correct pictorial for the pressure increase would be

A

Part C Using P1V1 = P2V2

with,

P1 = 625 mmHg = 0.82237 atm

V1 = 0.215 L

P2 = 1.3 atm

V2 = ?

we get,

final volume V2 = 0.82237 x 0.215/1.3 = 0.136 L = 136.01 ml

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