A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ? first: This is a problem a. missing variable b. change in conditions C. 22.414 L/1 mol d. density = molar mass/22.414 L e. partial P second: we need to solve for using the equation f. P g. P1 h. P2 i. V j. V1 (not V) k. V2 (not V) I. n m. n1 (not n) n. n2 (not n) o. T p. T1 (not T) q. T2 (not T) r. R= 8.205 x 10-2 L atm/K mol s. 1 atm = 760 torr t. Ptotal = P1 + P2 + .... u. PV = nRT v. P¡V1 P2V2 ------ ------ finally: The answer is v. 4.000 L w. 2.250 L x. 0.7500 L у. 1.333 L
A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ? first: This is a problem a. missing variable b. change in conditions C. 22.414 L/1 mol d. density = molar mass/22.414 L e. partial P second: we need to solve for using the equation f. P g. P1 h. P2 i. V j. V1 (not V) k. V2 (not V) I. n m. n1 (not n) n. n2 (not n) o. T p. T1 (not T) q. T2 (not T) r. R= 8.205 x 10-2 L atm/K mol s. 1 atm = 760 torr t. Ptotal = P1 + P2 + .... u. PV = nRT v. P¡V1 P2V2 ------ ------ finally: The answer is v. 4.000 L w. 2.250 L x. 0.7500 L у. 1.333 L
Introductory Chemistry: A Foundation
9th Edition
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Donald J. DeCoste
Chapter13: Gases
Section: Chapter Questions
Problem 18QAP
Related questions
Question
![A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ?
first: This is a
problem
a. missing variable
b. change in conditions
C. 22.414 L/1 mol
d. density = molar mass/22.414 L
e. partial P
second: we need to solve for
using the equation
f. P
g. P1
h. P2
i. V
j. V1 (not V)
k. V2 (not V)
I. n
m. n1 (not n)
n. n2 (not n)
О. Т
p. T1 (not T)
q. T2 (not T)
r. R= 8.205 x 10-2 L atm/K mol
S. 1 atm = 760 torr
t. Ptotal = P1 + P2 +
u. PV = nRT
v. P¡V1
P2V2
n2T2
finally: The answer is
V. 4.000 L
w. 2.250 L
X. 0.7500 L
у. 1.333 L](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3af619a8-05d7-4a08-b6ab-836abf944ec6%2F9991736e-0b11-42d8-8dda-6e6232b61385%2F66j38mt_processed.png&w=3840&q=75)
Transcribed Image Text:A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the new volume of the balloon ?
first: This is a
problem
a. missing variable
b. change in conditions
C. 22.414 L/1 mol
d. density = molar mass/22.414 L
e. partial P
second: we need to solve for
using the equation
f. P
g. P1
h. P2
i. V
j. V1 (not V)
k. V2 (not V)
I. n
m. n1 (not n)
n. n2 (not n)
О. Т
p. T1 (not T)
q. T2 (not T)
r. R= 8.205 x 10-2 L atm/K mol
S. 1 atm = 760 torr
t. Ptotal = P1 + P2 +
u. PV = nRT
v. P¡V1
P2V2
n2T2
finally: The answer is
V. 4.000 L
w. 2.250 L
X. 0.7500 L
у. 1.333 L
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Follow-up Question
![A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the
new volume of the balloon ?
first: This is a
a. missing variable
d. density = molar mass/22.414 L
second: we need to solve for
o. T
760 torr
f. P g. P₁ h. P₂ i. V j. V₁ (not V)
n. n₂ (not n)
p. T₁ (not T)
t. Ptotal = P₁ + P₂ +
u. PV = nRT
v. P₁V₁
P2V₂
n₁T₁
n₂T2
finally: The answer is
w. 0.7500 L
problem
b. change in conditions
x. 1.333 L
q. T₂ (not T)
e. partial P
y. 2.250 L
c. 22.414 L/1 mol
using the equation
k. V₂ (not V)
z. 4.000 L
I. n
r. R = 8.205 x 10-² L atm/K mol
m. n₁ (not n)
s. 1 atm =](https://content.bartleby.com/qna-images/question/0ca20f59-eb0e-436d-9fd7-0f9d240efea3/3f377612-99c8-4872-88cd-ade9991adaac/3t6iq8s_thumbnail.png)
Transcribed Image Text:A 3.000 L balloon is heated from 3000.0 K to 4000.0 K at a pressure of 1.900 atm. What is the
new volume of the balloon ?
first: This is a
a. missing variable
d. density = molar mass/22.414 L
second: we need to solve for
o. T
760 torr
f. P g. P₁ h. P₂ i. V j. V₁ (not V)
n. n₂ (not n)
p. T₁ (not T)
t. Ptotal = P₁ + P₂ +
u. PV = nRT
v. P₁V₁
P2V₂
n₁T₁
n₂T2
finally: The answer is
w. 0.7500 L
problem
b. change in conditions
x. 1.333 L
q. T₂ (not T)
e. partial P
y. 2.250 L
c. 22.414 L/1 mol
using the equation
k. V₂ (not V)
z. 4.000 L
I. n
r. R = 8.205 x 10-² L atm/K mol
m. n₁ (not n)
s. 1 atm =
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