Using the conversion between atmospheres and kPa (1 atm = 101.3 kPa), calculate the value for R with units of L-kPa-K-1-mol-1. (Show all work and round to three significant digits.)
Using the conversion between atmospheres and kPa (1 atm = 101.3 kPa), calculate the value for R with units of L-kPa-K-1-mol-1. (Show all work and round to three significant digits.)
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 Statement:**
Using the conversion between atmospheres and kPa (1 atm = 101.3 kPa), calculate the value for R with units of L·kPa·K⁻¹·mol⁻¹. (Show all work and round to three significant digits.)
---
**Solution Explanation:**
1. **Conversions and Constants:**
- 1 atm = 101.3 kPa
- Standard value of R in L·atm·K⁻¹·mol⁻¹ is 0.0821.
2. **Convert R from L·atm·K⁻¹·mol⁻¹ to L·kPa·K⁻¹·mol⁻¹:**
To convert, use the relation between atm and kPa:
\[
R = 0.0821 \, \text{L·atm·K}^{-1}\text{mol}^{−1} \times \frac{101.3 \, \text{kPa}}{1 \, \text{atm}}
\]
3. **Calculation:**
\[
R = 0.0821 \times 101.3 = 8.34 \, \text{L·kPa·K}^{-1}\text{mol}^{−1}
\]
4. **Rounded Value:**
\[
R = 8.34 \, \text{L·kPa·K}^{-1}\text{mol}^{−1} \quad (\text{rounded to three significant digits})
\]
---
**Conclusion:**
The calculated value of R with units of L·kPa·K⁻¹·mol⁻¹ is 8.34, which is obtained by converting the standard value of R from atm to kPa using the conversion factor of 1 atm = 101.3 kPa.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbbeeb476-a64f-4459-b8af-acf5dbdffec4%2F41438471-679e-48c3-8a35-399a35ba31f9%2Fiv90yqe_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Using the conversion between atmospheres and kPa (1 atm = 101.3 kPa), calculate the value for R with units of L·kPa·K⁻¹·mol⁻¹. (Show all work and round to three significant digits.)
---
**Solution Explanation:**
1. **Conversions and Constants:**
- 1 atm = 101.3 kPa
- Standard value of R in L·atm·K⁻¹·mol⁻¹ is 0.0821.
2. **Convert R from L·atm·K⁻¹·mol⁻¹ to L·kPa·K⁻¹·mol⁻¹:**
To convert, use the relation between atm and kPa:
\[
R = 0.0821 \, \text{L·atm·K}^{-1}\text{mol}^{−1} \times \frac{101.3 \, \text{kPa}}{1 \, \text{atm}}
\]
3. **Calculation:**
\[
R = 0.0821 \times 101.3 = 8.34 \, \text{L·kPa·K}^{-1}\text{mol}^{−1}
\]
4. **Rounded Value:**
\[
R = 8.34 \, \text{L·kPa·K}^{-1}\text{mol}^{−1} \quad (\text{rounded to three significant digits})
\]
---
**Conclusion:**
The calculated value of R with units of L·kPa·K⁻¹·mol⁻¹ is 8.34, which is obtained by converting the standard value of R from atm to kPa using the conversion factor of 1 atm = 101.3 kPa.
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