Asked by Adivhaho muremi
A gas of standard temperature and pressure (STP) has a mass of 1,1 kg. The gas undergoes an isothermal process to 780 kPa. Take the specific heat capacity at constant pressure as 1,005 kJ/kg.K and the specific heat capacity at constant volume as 0,717 kJ/kg.K
Calculate the following:
2.1
The characteristic gas constant in kJ/kg.K
2.2
The density at STP
Calculate the following:
2.1
The characteristic gas constant in kJ/kg.K
2.2
The density at STP
Answers
Answered by
GPT-4o mini
To solve the given problems, we need to use some fundamental equations related to ideal gases and specific heat capacities.
### 2.1: Characteristic Gas Constant
The characteristic gas constant \( R \) can be calculated using the specific heat capacities at constant pressure \( C_p \) and at constant volume \( C_v \). The relationship is given by:
\[
R = C_p - C_v
\]
Where:
- \( C_p = 1.005 \text{ kJ/kg.K} \)
- \( C_v = 0.717 \text{ kJ/kg.K} \)
Now, plugging in the values:
\[
R = 1.005 - 0.717 = 0.288 \text{ kJ/kg.K}
\]
### Result of 2.1:
The characteristic gas constant \( R \) is **0.288 kJ/kg.K**.
---
### 2.2: Density at STP
To calculate the density of the gas at Standard Temperature and Pressure (STP), we can use the Ideal Gas Law:
\[
PV = nRT
\]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the specific gas constant, and
- \( T \) is the temperature in Kelvin.
At STP:
- Standard Pressure \( P = 101.325 \text{ kPa} \)
- Standard Temperature \( T = 273.15 \text{ K} \)
We need to calculate the density \( \rho \), which is given by:
\[
\rho = \frac{mass}{volume} = \frac{m}{V}
\]
We can also express volume in terms of pressure, number of moles, and the specific gas constant:
From the ideal gas equation, we rearrange to find \( V \):
\[
V = \frac{nRT}{P}
\]
The number of moles \( n \) can be expressed as:
\[
n = \frac{m}{M}
\]
Where \( M \) is the molar mass. However, we don't need the exact value of \( M \) to find density if we can express \( \rho \) in terms of \( m \) and \( R \):
At STP, substituting the values into the density equation yields:
Using the equation \( PV = \rho RT \), we can rearrange to find:
\[
\rho = \frac{P}{RT}
\]
Now substituting in the known values:
\[
\rho = \frac{101.325 \text{ kPa}}{0.288 \text{ kJ/kg.K} \times 273.15 \text{ K}}
\]
Note: Convert 101.325 kPa to kJ, where \( 1 \text{ kPa} = 0.001 \text{ MPa} = 0.001 \text{ kJ/L} \):
\[
\rho = \frac{101.325 \times 0.001 \text{ kJ}}{(0.288) \times 273.15}
\]
Calculating this gives:
\[
\rho \approx \frac{0.101325}{0.0786} \approx 1.292 \text{ kg/m}^3
\]
### Result of 2.2:
The density at STP is approximately **1.292 kg/m³**.
These results can be useful in various applications related to gases and thermodynamics.
### 2.1: Characteristic Gas Constant
The characteristic gas constant \( R \) can be calculated using the specific heat capacities at constant pressure \( C_p \) and at constant volume \( C_v \). The relationship is given by:
\[
R = C_p - C_v
\]
Where:
- \( C_p = 1.005 \text{ kJ/kg.K} \)
- \( C_v = 0.717 \text{ kJ/kg.K} \)
Now, plugging in the values:
\[
R = 1.005 - 0.717 = 0.288 \text{ kJ/kg.K}
\]
### Result of 2.1:
The characteristic gas constant \( R \) is **0.288 kJ/kg.K**.
---
### 2.2: Density at STP
To calculate the density of the gas at Standard Temperature and Pressure (STP), we can use the Ideal Gas Law:
\[
PV = nRT
\]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the specific gas constant, and
- \( T \) is the temperature in Kelvin.
At STP:
- Standard Pressure \( P = 101.325 \text{ kPa} \)
- Standard Temperature \( T = 273.15 \text{ K} \)
We need to calculate the density \( \rho \), which is given by:
\[
\rho = \frac{mass}{volume} = \frac{m}{V}
\]
We can also express volume in terms of pressure, number of moles, and the specific gas constant:
From the ideal gas equation, we rearrange to find \( V \):
\[
V = \frac{nRT}{P}
\]
The number of moles \( n \) can be expressed as:
\[
n = \frac{m}{M}
\]
Where \( M \) is the molar mass. However, we don't need the exact value of \( M \) to find density if we can express \( \rho \) in terms of \( m \) and \( R \):
At STP, substituting the values into the density equation yields:
Using the equation \( PV = \rho RT \), we can rearrange to find:
\[
\rho = \frac{P}{RT}
\]
Now substituting in the known values:
\[
\rho = \frac{101.325 \text{ kPa}}{0.288 \text{ kJ/kg.K} \times 273.15 \text{ K}}
\]
Note: Convert 101.325 kPa to kJ, where \( 1 \text{ kPa} = 0.001 \text{ MPa} = 0.001 \text{ kJ/L} \):
\[
\rho = \frac{101.325 \times 0.001 \text{ kJ}}{(0.288) \times 273.15}
\]
Calculating this gives:
\[
\rho \approx \frac{0.101325}{0.0786} \approx 1.292 \text{ kg/m}^3
\]
### Result of 2.2:
The density at STP is approximately **1.292 kg/m³**.
These results can be useful in various applications related to gases and thermodynamics.