Question

An isolated quantity of an ideal gas at 292.9 K has a volume of 22.67 L at a pressure of 1.06 atm. What is the volume of this gas sample when the absolute temperature is reduced to one third and the pressure is divided by four?

Answer 1

Initial                                                             final

T1   = 292.9K                                              T2   = 292.9/3    = 97.64K

V1   = 22.67L                                               V2 =

P1   = 1.06atm                                             P2 = 1.06/4 = 0.265atm

              P1V1/T1             =       P2V2/T2

                    V2                =      P1V1T2/T1P2

                                        =     1.06*22.67*97.64/(292.9*0.265)

                                       =    30.23 L