Let's solve the problem step-by-step using the principles of the ideal gas law.
The ideal gas law is expressed as:
[tex]\[ PV = nRT \][/tex]
Where:
- [tex]\( P \)[/tex] is the pressure of the gas.
- [tex]\( V \)[/tex] is the volume of the gas.
- [tex]\( n \)[/tex] is the number of moles of the gas.
- [tex]\( R \)[/tex] is the gas constant.
- [tex]\( T \)[/tex] is the absolute temperature of the gas.
Given:
- The initial state of the gas has pressure [tex]\( P_{\text{initial}} \)[/tex] and temperature [tex]\( T_{\text{initial}} \)[/tex].
- Both the temperature and pressure are tripled.
So, for the new state:
- The new pressure, [tex]\( P_{\text{new}} \)[/tex], is [tex]\( 3P_{\text{initial}} \)[/tex].
- The new temperature, [tex]\( T_{\text{new}} \)[/tex], is [tex]\( 3T_{\text{initial}} \)[/tex].
Let's write the ideal gas law for both the initial state and the new state.
Initial state:
[tex]\[ P_{\text{initial}} V_{\text{initial}} = nR T_{\text{initial}} \][/tex]
New state:
[tex]\[ P_{\text{new}} V_{\text{new}} = nR T_{\text{new}} \][/tex]
Substitute the values for the new pressure and temperature:
[tex]\[ (3P_{\text{initial}}) V_{\text{new}} = nR (3T_{\text{initial}}) \][/tex]
We can simplify this equation by dividing both sides by 3:
[tex]\[ P_{\text{initial}} V_{\text{new}} = nR T_{\text{initial}} \][/tex]
Notice that this simplifies to the initial state equation:
[tex]\[ P_{\text{initial}} V_{\text{initial}} = nR T_{\text{initial}} \][/tex]
This shows that:
[tex]\[ V_{\text{initial}} = V_{\text{new}} \][/tex]
Therefore, the ratio of the new volume to the initial volume is:
[tex]\[ \frac{V_{\text{new}}}{V_{\text{initial}}} = \frac{V_{\text{initial}}}{V_{\text{initial}}} = 1 \][/tex]
So, the ratio of the new volume to the initial volume is 1:1.
