Question
How does doubling both the volume and pressure of a mass of Nitrogen initially at 0°C affect its thermodynamic properties?
Answers
GPT-4o mini
To understand the effects of doubling both the volume and pressure of a mass of nitrogen initially at 0°C (273.15 K), we need to analyze the situation using the ideal gas law and principles from thermodynamics.
### Initial Conditions
Let's denote the initial conditions of nitrogen:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( n \) = number of moles of nitrogen
- \( R \) = ideal gas constant
- \( T_1 \) = 273.15 K (0°C)
From the ideal gas equation:
\[
P_1 V_1 = n R T_1
\]
### Final Conditions
According to the problem, if the volume and pressure are both doubled, then:
- \( P_2 = 2P_1 \)
- \( V_2 = 2V_1 \)
### Applying the Ideal Gas Law to Final Conditions
Using the ideal gas law again for the final conditions:
\[
P_2 V_2 = n R T_2
\]
Substituting the new values into this equation:
\[
(2P_1)(2V_1) = n R T_2
\]
\[
4P_1 V_1 = n R T_2
\]
### Relating Initial and Final States
Now, we can relate the final volume-pressure change back to the initial conditions. Since \( P_1 V_1 = n R T_1 \), we have:
\[
4(P_1 V_1) = n R T_2 \implies 4(n R T_1) = n R T_2
\]
Simplifying this, we divide both sides by \( n R \):
\[
4T_1 = T_2
\]
Given \( T_1 = 273.15 \) K (0°C):
\[
T_2 = 4 \times 273.15 = 1092.6 \text{ K}
\]
### Summary of Changes in Thermodynamic Properties
1. **Temperature Increase**: The final temperature \( T_2 \) of the nitrogen increases to 1092.6 K due to the increase in both pressure and volume.
2. **No Change in Number of Moles**: The number of moles \( n \) remains constant as the gas system is closed.
3. **Pressure-Volume Relationship**: While both pressure and volume are increased, this leads to a significant increase in temperature.
Overall, doubling both the volume and pressure of a gas at a constant amount of moles clearly results in a substantial increase in its temperature according to the ideal gas laws. This exemplifies how changes to pressure and volume influence thermodynamic properties like temperature.
### Initial Conditions
Let's denote the initial conditions of nitrogen:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( n \) = number of moles of nitrogen
- \( R \) = ideal gas constant
- \( T_1 \) = 273.15 K (0°C)
From the ideal gas equation:
\[
P_1 V_1 = n R T_1
\]
### Final Conditions
According to the problem, if the volume and pressure are both doubled, then:
- \( P_2 = 2P_1 \)
- \( V_2 = 2V_1 \)
### Applying the Ideal Gas Law to Final Conditions
Using the ideal gas law again for the final conditions:
\[
P_2 V_2 = n R T_2
\]
Substituting the new values into this equation:
\[
(2P_1)(2V_1) = n R T_2
\]
\[
4P_1 V_1 = n R T_2
\]
### Relating Initial and Final States
Now, we can relate the final volume-pressure change back to the initial conditions. Since \( P_1 V_1 = n R T_1 \), we have:
\[
4(P_1 V_1) = n R T_2 \implies 4(n R T_1) = n R T_2
\]
Simplifying this, we divide both sides by \( n R \):
\[
4T_1 = T_2
\]
Given \( T_1 = 273.15 \) K (0°C):
\[
T_2 = 4 \times 273.15 = 1092.6 \text{ K}
\]
### Summary of Changes in Thermodynamic Properties
1. **Temperature Increase**: The final temperature \( T_2 \) of the nitrogen increases to 1092.6 K due to the increase in both pressure and volume.
2. **No Change in Number of Moles**: The number of moles \( n \) remains constant as the gas system is closed.
3. **Pressure-Volume Relationship**: While both pressure and volume are increased, this leads to a significant increase in temperature.
Overall, doubling both the volume and pressure of a gas at a constant amount of moles clearly results in a substantial increase in its temperature according to the ideal gas laws. This exemplifies how changes to pressure and volume influence thermodynamic properties like temperature.