2 kg of gas occupying 0,7 m had on original temperature of 15 °C. It was then heated at constant volume until its temperature become 135 °C. How much heat was transferred to the gas and what was its final pressure? Take Cv = 0.72 kl/ kg.K and R = 0.29 kJ/ kg.K
2 kg of gas occupying 0,7 m had on original temperature of 15 °C. It was then heated at constant volume until its temperature become 135 °C. How much heat was transferred to the gas and what was its final pressure? Take Cv = 0.72 kl/ kg.K and R = 0.29 kJ/ kg.K
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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![### Example 2
A gas system problem is described below, and we will solve it step by step for educational purposes.
**Problem Statement:**
2 kilograms of gas occupying 0.7 cubic meters had an original temperature of 15°C. It was then heated at a constant volume until its temperature reached 135°C. We need to determine how much heat was transferred to the gas and what its final pressure was. For the calculations, take the specific heat at constant volume (Cv) as 0.72 kJ/kg·K and the gas constant (R) as 0.29 kJ/kg·K.
#### Solution:
To solve this problem, we need to apply the concepts of thermodynamics, particularly focusing on the first law of thermodynamics and the ideal gas law.
1. **Heat Transfer Calculation:**
The heat transferred (Q) to the gas at constant volume can be calculated using the formula:
\[
Q = m \cdot C_v \cdot \Delta T
\]
where:
- \(m\) is the mass of the gas (2 kg)
- \(C_v\) is the specific heat at constant volume (0.72 kJ/kg·K)
- \(\Delta T\) is the change in temperature
First, convert temperatures from Celsius to Kelvin:
\[
T_1 = 15°C + 273.15 = 288.15 K
\]
\[
T_2 = 135°C + 273.15 = 408.15 K
\]
Now, calculate the change in temperature:
\[
\Delta T = T_2 - T_1 = 408.15 K - 288.15 K = 120 K
\]
Plug these values into the formula:
\[
Q = 2 \, \text{kg} \times 0.72 \, \text{kJ/kg·K} \times 120 \, \text{K} = 172.8 \, \text{kJ}
\]
Therefore, the heat transferred to the gas is \(172.8 \, \text{kJ}\).
2. **Final Pressure Calculation:**
The final pressure can be determined using the ideal gas law:
\[
PV = nRT](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcd526135-39bd-4cf8-a87a-c8fb9db939a2%2F4352f500-302b-4a5d-afe5-fb65f90e6fbb%2Fez1bo2i_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Example 2
A gas system problem is described below, and we will solve it step by step for educational purposes.
**Problem Statement:**
2 kilograms of gas occupying 0.7 cubic meters had an original temperature of 15°C. It was then heated at a constant volume until its temperature reached 135°C. We need to determine how much heat was transferred to the gas and what its final pressure was. For the calculations, take the specific heat at constant volume (Cv) as 0.72 kJ/kg·K and the gas constant (R) as 0.29 kJ/kg·K.
#### Solution:
To solve this problem, we need to apply the concepts of thermodynamics, particularly focusing on the first law of thermodynamics and the ideal gas law.
1. **Heat Transfer Calculation:**
The heat transferred (Q) to the gas at constant volume can be calculated using the formula:
\[
Q = m \cdot C_v \cdot \Delta T
\]
where:
- \(m\) is the mass of the gas (2 kg)
- \(C_v\) is the specific heat at constant volume (0.72 kJ/kg·K)
- \(\Delta T\) is the change in temperature
First, convert temperatures from Celsius to Kelvin:
\[
T_1 = 15°C + 273.15 = 288.15 K
\]
\[
T_2 = 135°C + 273.15 = 408.15 K
\]
Now, calculate the change in temperature:
\[
\Delta T = T_2 - T_1 = 408.15 K - 288.15 K = 120 K
\]
Plug these values into the formula:
\[
Q = 2 \, \text{kg} \times 0.72 \, \text{kJ/kg·K} \times 120 \, \text{K} = 172.8 \, \text{kJ}
\]
Therefore, the heat transferred to the gas is \(172.8 \, \text{kJ}\).
2. **Final Pressure Calculation:**
The final pressure can be determined using the ideal gas law:
\[
PV = nRT
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