Answer:
The temperature of the methane gas will drop to 22.127 °C.
Explanation:
To solve the question, we note tht the combined gas equation is given by
[tex]\frac{P_1\cdot V_1}{T_1} = \frac{P_2\cdot V_2}{T_2}[/tex]
Where:
P₁ = Initial pressure = 98.6 kPa
V₁ = Initial volume = 1000 mL
T₁ = Initial temperature = 25°c = 298.15 K
P₂ = Final pressure = 108.5 kPa
V₂ = Final volume = 900 mL
T₂ = Final temperature = Required
Therefore we have
[tex]T_2= \frac{P_2\cdot V_2\cdot T_1}{P_1\cdot V_1}[/tex] = [tex]\\\frac{108.5\cdot 900\cdot 298.15}{98.6\cdot 1000}[/tex] = 295.277 K
The temperature T₂ < T₁
Therefore what will happen is that the temperature of the methane gas will drop by 2.873 °C from 25 °C to 22.127 °C.
