The volume of a fixed amount of gas is doubled, and the absolute temperature is held constant. According to Boyle's law, how has the pressure of the gas changed?

A. It has increased to two times its original value.
B. It has increased to four times its original value.
C. It has decreased to one-half its original value.
D. It has stayed the same.



Answer :

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.