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 8915 J. What is the specific heat of the gas? C= I Enter numeric value 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 8915 J. What is the specific heat of the gas? C= I Enter numeric value 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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![**Problem Statement:**
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 8915 J. What is the specific heat of the gas?
**Solution:**
To find the specific heat capacity (\(c\)) of the gas, we use the formula relating heat (\(q\)), mass (\(m\)), specific heat capacity (\(c\)), and change in temperature (\(\Delta T\)):
\[
q = mc\Delta T
\]
The change in internal energy (\(\Delta U\)) relates to heat added and work done:
\[
\Delta U = q + W
\]
Where:
- \(\Delta U = 8915 \, \text{J}\) (change in internal energy)
- \(W = -346 \, \text{J}\) (work done by the system, negative because work is done by the gas)
From the equation \(\Delta U = q + W\), solve for \(q\):
\[
q = \Delta U - W = 8915 \, \text{J} + 346 \, \text{J} = 9261 \, \text{J}
\]
Now substitute \(q\) into the specific heat formula:
\[
9261 \, \text{J} = 80.0 \, \text{g} \cdot c \cdot (225 - 25) \, \text{°C}
\]
\[
9261 = 80.0 \cdot c \cdot 200
\]
Solve for \(c\):
\[
c = \frac{9261}{80.0 \cdot 200} = \frac{9261}{16000} \approx 0.579 \, \text{J/(g·°C)}
\]
Therefore, the specific heat capacity of the gas is approximately \(0.579 \, \text{J/(g·°C)}\).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F76884ef0-0050-4a4d-a83f-5c6d84fc3fcd%2F15596826-c087-414a-84e9-b6faaf59cb18%2Fyifo4ix_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
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 8915 J. What is the specific heat of the gas?
**Solution:**
To find the specific heat capacity (\(c\)) of the gas, we use the formula relating heat (\(q\)), mass (\(m\)), specific heat capacity (\(c\)), and change in temperature (\(\Delta T\)):
\[
q = mc\Delta T
\]
The change in internal energy (\(\Delta U\)) relates to heat added and work done:
\[
\Delta U = q + W
\]
Where:
- \(\Delta U = 8915 \, \text{J}\) (change in internal energy)
- \(W = -346 \, \text{J}\) (work done by the system, negative because work is done by the gas)
From the equation \(\Delta U = q + W\), solve for \(q\):
\[
q = \Delta U - W = 8915 \, \text{J} + 346 \, \text{J} = 9261 \, \text{J}
\]
Now substitute \(q\) into the specific heat formula:
\[
9261 \, \text{J} = 80.0 \, \text{g} \cdot c \cdot (225 - 25) \, \text{°C}
\]
\[
9261 = 80.0 \cdot c \cdot 200
\]
Solve for \(c\):
\[
c = \frac{9261}{80.0 \cdot 200} = \frac{9261}{16000} \approx 0.579 \, \text{J/(g·°C)}
\]
Therefore, the specific heat capacity of the gas is approximately \(0.579 \, \text{J/(g·°C)}\).
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