ScoresA 150.0-mL bubble of hot gases at 209 °C and 2.00 atm escapes from an active volcano.eTextPart ADocument SharingWhat is the new volume of the bubble outside the volcano where the temperature is -31 °C and the pressure is 0.80 atm?User SettingsExpress your answer to two significant figures and include the appropriate units.Course ToolsμΑValueV =UnitsSubmitRequest AnswerP PearsonCopyright O 2019 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy PolicyPermissions/Contact UsTREL DCtVCy TOTT e eTTocsWomen'sOnly 2 left in stock - order soon.Only 10 left in stoc30,639OCT128WescF1F2F3F4F5F6F7F8F9FTOFU%&2735QWTtA4

Question
Asked Oct 29, 2019
Scores
A 150.0-mL bubble of hot gases at 209 °C and 2.00 atm escapes from an active volcano.
eText
Part A
Document Sharing
What is the new volume of the bubble outside the volcano where the temperature is -31 °C and the pressure is 0.80 atm?
User Settings
Express your answer to two significant figures and include the appropriate units.
Course Tools
μΑ
Value
V =
Units
Submit
Request Answer
P Pearson
Copyright O 2019 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy
Permissions
/
Contact Us
TREL DCtVCy TOTT e eTTocs
Women's
Only 2 left in stock - order soon.
Only 10 left in stoc
30,639
OCT
1
28
W
esc
F1
F2
F3
F4
F5
F6
F7
F8
F9
FTO
FU
%
&
2
7
3
5
Q
W
T
tA4
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Scores A 150.0-mL bubble of hot gases at 209 °C and 2.00 atm escapes from an active volcano. eText Part A Document Sharing What is the new volume of the bubble outside the volcano where the temperature is -31 °C and the pressure is 0.80 atm? User Settings Express your answer to two significant figures and include the appropriate units. Course Tools μΑ Value V = Units Submit Request Answer P Pearson Copyright O 2019 Pearson Education Inc. All rights reserved. | Terms of Use | Privacy Policy Permissions / Contact Us TREL DCtVCy TOTT e eTTocs Women's Only 2 left in stock - order soon. Only 10 left in stoc 30,639 OCT 1 28 W esc F1 F2 F3 F4 F5 F6 F7 F8 F9 FTO FU % & 2 7 3 5 Q W T tA4

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check_circleExpert Solution
Step 1

Given:

Initial volume of Bubble =  150 mL = V1

Initial T = 209 o C = 482.15 K = T1

Initial P = 2 atm = P1

Final volume =? = V2

Final T = -31 o C = 242.15 K = T2

Final P = 0.80 atm = P2

Step 2

Calculation:

From the ideal gas equation,

(P1V1)/T1=(P2V2)/T2

Substitute the values known and simplify,

(2 X 150 ) / 482.15  = (0.80 X  V2 )/ 242.15

300 / 482.15  = (0.80 X  V2 )/ 242.15

0...

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