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
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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![### Problem:
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?
### Step-by-Step Solution:
#### First:
Identify the type of problem:
**This is a:**
- [ ] a. missing variable
- [x] b. change in conditions
- [ ] c. 22.414 L/1 mol
- [ ] d. density = molar mass/22.414 L
- [ ] e. partial P
#### Second:
Determine what needs solving:
**We need to solve for:**
- [x] k. V₂ (not V)
**Using the equation:**
- v. \[ \frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2} \]
### Information and Formulas:
- **Symbols and values:**
- f. P
- g. P₁
- h. P₂
- i. V
- j. V₁ (not V)
- l. 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
- **Total Pressure:**
- t. \( P_{\text{total}} = P_1 + P_2 + \ldots \)
- **Ideal Gas Law:**
- u. PV = nRT
### Finally:
Select the correct answer for the new volume:
- v. \[4.000 L\]
- w. \[2.250 L\]
- x. \[0.7500 L\]
- y. **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:### Problem:
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?
### Step-by-Step Solution:
#### First:
Identify the type of problem:
**This is a:**
- [ ] a. missing variable
- [x] b. change in conditions
- [ ] c. 22.414 L/1 mol
- [ ] d. density = molar mass/22.414 L
- [ ] e. partial P
#### Second:
Determine what needs solving:
**We need to solve for:**
- [x] k. V₂ (not V)
**Using the equation:**
- v. \[ \frac{P_1V_1}{n_1T_1} = \frac{P_2V_2}{n_2T_2} \]
### Information and Formulas:
- **Symbols and values:**
- f. P
- g. P₁
- h. P₂
- i. V
- j. V₁ (not V)
- l. 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
- **Total Pressure:**
- t. \( P_{\text{total}} = P_1 + P_2 + \ldots \)
- **Ideal Gas Law:**
- u. PV = nRT
### Finally:
Select the correct answer for the new volume:
- v. \[4.000 L\]
- w. \[2.250 L\]
- x. \[0.7500 L\]
- y. **1.333 L**
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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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