Answer:
The temperature will change and become 2/3 of its original.
Explanation:
Using Ideal gas equation for same mole of gas as
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
Given ,
The volume of the sample gets reduced 1/3 of the original. So,
V₂ = 1/3V₁
The pressure of the sample is doubled of the original. So,
P₂ = 2P₁
Using above equation as:
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{P_2}\times {V_2}}{T_2}[/tex]
[tex] \frac {{P_1}\times {V_1}}{T_1}=\frac {{2\times P_1}\times {\frac{1}{3}\times V_1}}{T_2}[/tex]
[tex]T_2=\frac{2}{3}\times T_1[/tex]
The temperature will change and become 2/3 of its original.
