Question

In: Chemistry

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.

Solutions

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


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