Question
The absolute temperature of a sample of monatomic ideal gas is doubled at constant volume. What effect, if any, does this have on the pressure and density of the sample of gas?
Pressure Density
1. Doubles
Remains the same
2. Remains the same
Doubles
3. Doubles
Multiplied by 4
4. Multiplied by 4
Doubles
5. Remains the same
Remains the same
Pressure Density
1. Doubles
Remains the same
2. Remains the same
Doubles
3. Doubles
Multiplied by 4
4. Multiplied by 4
Doubles
5. Remains the same
Remains the same
Answers
Answered by
GPT-4o mini
To analyze the effect of doubling the absolute temperature on the pressure and density of a monatomic ideal gas at constant volume, we can make use of the ideal gas law, which is given by:
\[
PV = nRT
\]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant, and
- \( T \) is the absolute temperature.
Since the volume \( V \) remains constant in this scenario, we can study the relationship between pressure and temperature:
From the ideal gas law, we can express pressure as:
\[
P = \frac{nRT}{V}
\]
As the temperature \( T \) is doubled (let's denote the initial temperature as \( T \) and the new temperature as \( 2T \)), we can substitute \( 2T \) into the equation for pressure:
\[
P_{\text{new}} = \frac{nR(2T)}{V} = 2 \left( \frac{nRT}{V} \right) = 2P_{\text{initial}}
\]
Thus, the pressure of the gas will double when the absolute temperature is doubled at constant volume.
Now, let's analyze the effect on density. The density \( \rho \) of the gas is given by:
\[
\rho = \frac{m}{V}
\]
Where \( m \) is the mass of the gas. Since we are dealing with an ideal gas, we can relate density to the number of moles \( n \) and molar mass \( M \):
\[
\rho = \frac{nM}{V}
\]
The number of moles \( n \) is a function of the temperature, volume, and pressure, but at constant volume and with an ideal gas, density is mostly influenced by the number of particles (and thus moles) present. Since the pressure increases while volume is constant, the density itself is not directly influenced by the temperature change unless additional constraints apply (like changing the mass of the gas).
Here, since the mass \( m \) of the gas does not change and the volume \( V \) remains constant, the density \( \rho \) remains unchanged.
To summarize:
- Pressure doubles.
- Density remains the same.
Thus, the correct answer is:
**1. Doubles; Remains the same.**
\[
PV = nRT
\]
Where:
- \( P \) is the pressure,
- \( V \) is the volume,
- \( n \) is the number of moles,
- \( R \) is the ideal gas constant, and
- \( T \) is the absolute temperature.
Since the volume \( V \) remains constant in this scenario, we can study the relationship between pressure and temperature:
From the ideal gas law, we can express pressure as:
\[
P = \frac{nRT}{V}
\]
As the temperature \( T \) is doubled (let's denote the initial temperature as \( T \) and the new temperature as \( 2T \)), we can substitute \( 2T \) into the equation for pressure:
\[
P_{\text{new}} = \frac{nR(2T)}{V} = 2 \left( \frac{nRT}{V} \right) = 2P_{\text{initial}}
\]
Thus, the pressure of the gas will double when the absolute temperature is doubled at constant volume.
Now, let's analyze the effect on density. The density \( \rho \) of the gas is given by:
\[
\rho = \frac{m}{V}
\]
Where \( m \) is the mass of the gas. Since we are dealing with an ideal gas, we can relate density to the number of moles \( n \) and molar mass \( M \):
\[
\rho = \frac{nM}{V}
\]
The number of moles \( n \) is a function of the temperature, volume, and pressure, but at constant volume and with an ideal gas, density is mostly influenced by the number of particles (and thus moles) present. Since the pressure increases while volume is constant, the density itself is not directly influenced by the temperature change unless additional constraints apply (like changing the mass of the gas).
Here, since the mass \( m \) of the gas does not change and the volume \( V \) remains constant, the density \( \rho \) remains unchanged.
To summarize:
- Pressure doubles.
- Density remains the same.
Thus, the correct answer is:
**1. Doubles; Remains the same.**