An 80.0 g sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 6285 J. What is the specific heat of the gas? D C = J/(g. °C)
An 80.0 g sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 6285 J. What is the specific heat of the gas? D C = J/(g. °C)
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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![**Question:**
An 80.0 g sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 6285 J. What is the specific heat of the gas?
**Solution:**
To find the specific heat of the gas (\(c\)), we can use the formula derived from the first law of thermodynamics:
\[
\Delta U = q - w
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(q\) is the heat added to the system,
- \(w\) is the work done by the system.
Given:
- \(\Delta U = 6285 \, \text{J}\),
- \(w = 346 \, \text{J}\).
Rearranging the formula to find \(q\):
\[
q = \Delta U + w = 6285 \, \text{J} + 346 \, \text{J} = 6631 \, \text{J}
\]
Next, we use the formula for calculating specific heat:
\[
q = mc\Delta T
\]
Solving for \(c\):
\[
c = \frac{q}{m \Delta T}
\]
Where:
- \(m = 80.0 \, \text{g}\),
- \(\Delta T = 225 \, °C - 25 \, °C = 200 \, °C\),
- \(q = 6631 \, \text{J}\).
Plug in the values:
\[
c = \frac{6631 \, \text{J}}{80.0 \, \text{g} \times 200 \, °C}
\]
\[
c = \frac{6631}{16000} \, \text{J/g°C}
\]
\[
c = 0.414 \, \text{J/g°C}
\]
The specific heat of the gas is 0.414 J/g°C.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffdc195bb-9009-47db-866e-65b6514806e8%2F3b0c91a5-f84d-4d75-9607-4517a8d46db0%2Ffmxm4xa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Question:**
An 80.0 g sample of a gas was heated from 25 °C to 225 °C. During this process, 346 J of work was done by the system and its internal energy increased by 6285 J. What is the specific heat of the gas?
**Solution:**
To find the specific heat of the gas (\(c\)), we can use the formula derived from the first law of thermodynamics:
\[
\Delta U = q - w
\]
Where:
- \(\Delta U\) is the change in internal energy,
- \(q\) is the heat added to the system,
- \(w\) is the work done by the system.
Given:
- \(\Delta U = 6285 \, \text{J}\),
- \(w = 346 \, \text{J}\).
Rearranging the formula to find \(q\):
\[
q = \Delta U + w = 6285 \, \text{J} + 346 \, \text{J} = 6631 \, \text{J}
\]
Next, we use the formula for calculating specific heat:
\[
q = mc\Delta T
\]
Solving for \(c\):
\[
c = \frac{q}{m \Delta T}
\]
Where:
- \(m = 80.0 \, \text{g}\),
- \(\Delta T = 225 \, °C - 25 \, °C = 200 \, °C\),
- \(q = 6631 \, \text{J}\).
Plug in the values:
\[
c = \frac{6631 \, \text{J}}{80.0 \, \text{g} \times 200 \, °C}
\]
\[
c = \frac{6631}{16000} \, \text{J/g°C}
\]
\[
c = 0.414 \, \text{J/g°C}
\]
The specific heat of the gas is 0.414 J/g°C.
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