Answer:
a)The temperature of the gas in increased by four times its original value.
Explanation:
For a fixed amount of a gas, the ideal gas equation can be written as follows:
[tex]\frac{pV}{T}=const.[/tex]
where
p is the gas pressure
V is the gas volume
T is the gas temperature (in Kelvins)
For a gas under transformation, the equation can also be rewritten as
[tex]\frac{p_1V_1}{T_1}=\frac{p_2V_2}{T_2}[/tex]
where the labels 1,2 refer to the initial and final conditions of the gas.
In this problem, we have:
- The pressure of the gas is doubled: [tex]p_2 = 2p_1[/tex]
- The volume of the gas is doubled: [tex]V_2=2V_1[/tex]
Substituting into the equation, we find what happens to the temperature:
[tex]\frac{p_1V_1}{T_1}=\frac{(2 p_1)(2V_1)}{T_2}[/tex]
[tex]\frac{1}{T_1}=\frac{4}{T_2}[/tex]
[tex]T_2 = 4T_1[/tex]
So, the correct choice is
a)The temperature of the gas in increased by four times its original value.
