To determine how the pressure of a gas changes when its volume and temperature are both doubled, we can use the ideal gas law, which states:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure
- [tex]\( V \)[/tex] is the volume
- [tex]\( n \)[/tex] is the number of moles of gas
- [tex]\( R \)[/tex] is the ideal gas constant
- [tex]\( T \)[/tex] is the absolute temperature
For a fixed amount of gas ([tex]\( n \)[/tex] and [tex]\( R \)[/tex] are constant), the relationship can be expressed as:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \][/tex]
Given the initial conditions:
- [tex]\( P_1 \)[/tex] is the initial pressure
- [tex]\( V_1 \)[/tex] is the initial volume
- [tex]\( T_1 \)[/tex] is the initial temperature
And the new conditions:
- The volume [tex]\( V_2 = 2V_1 \)[/tex] (the volume is doubled)
- The temperature [tex]\( T_2 = 2T_1 \)[/tex] (the absolute temperature is doubled)
We need to solve for the new pressure [tex]\( P_2 \)[/tex].
Substitute the given information into the equation:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{P_2 (2V_1)}{2T_1} \][/tex]
Simplify the equation:
[tex]\[ \frac{P_1 V_1}{T_1} = \frac{2V_1P_2}{2T_1} \][/tex]
[tex]\[ P_1 V_1 = P_2 V_1 \][/tex]
[tex]\[ P_1 = P_2 \][/tex]
From this, we can see that the new pressure [tex]\( P_2 \)[/tex] is equal to the initial pressure [tex]\( P_1 \)[/tex]. Therefore, the pressure of the gas has stayed the same.
The correct answer is:
- It has stayed the same.
