A sealed container holding 0.0255 L of an ideal gas at 0.993 atm and 67 °C...

A sealed container holding 0.0255 L of an ideal gas at 0.993 atm and 67 °C is placed into a refrigerator and cooled to 45 °C with no change in volume. Calculate the final pressure of the gas.

Expert Solution

Intila final

V1 = 0.0255L V2 = 0.0255L

P1 = 0.993atm P2 =

T1 = 67⁰C = 67+ 273 = 340K T2 = 45⁰C = 45+273 = 318K

P1V1/T1 = P2V2/T2

P2 = P1V1T2/T1V2

= 0.993*0.0255*318/340*0.0255

= 0.928atm >>>>>answer

queen_honey_blossom answered 1 year ago

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