Answer: -
If a gas at 250 K has a pressure of 0.00008 Pa and the pressure drops to 0.00002 Pa, 62.5 K is the new temperature
Explanation: -
Initial Pressure P₁ = 0.00008 Pa
Final Pressure P₂ = 0.00002 Pa
Initial Temperature T₁ = 250 K
Since the volume is not mentioned, it remains constant.
So initial Volume V₁ = Final volume V₂
Now using the combined gas equation
[tex] \frac{P1 V1}{T1} [/tex] = [tex] \frac{P2 V2}{T2} [/tex]
Final Temperature T₂ = [tex] \frac{P 2 V2 T1}{P1 V1} [/tex]
Plugging in the values,
T₂ = [tex] \frac{0.00002 Pa x V2 x 250 K}{0.00008 Pa x V1} [/tex]
As V₂ = V₁, they cancel each other.
T₂ = 62.5 K
