Question

1.A sample of krypton gas occupies a volume of 7.55 L at 54.0°C and0.530 atm.

If it is desired to decrease the volume of the gas sample to 4.69 L, whileincreasing its pressure to 0.710 atm, the temperature of the gas sample at the new volume and pressure must be °C.

2.

A sample of krypton gas occupies a volume of 6.09 L at 64.0°C and 360.torr.

If the volume of the gas sample is decreased to 3.81 L, while its temperature is decreased to 6.0°C, the resulting gas pressure will be   to rr.

3,

A helium-filled weather balloon has a volume of 849 L at 21.9°C and 757mmHg. It is released and rises to an altitude of 4.68 km, where the pressure is 501 mmHg and the temperature is –8.1°C.

The volume of the balloon at this altitude is L.

Answer 1

1)

                         Initial conditions                                                    Final conditions

Pressure = P1= 0.530 atm    Pressure = P2= 0.710 atm                                    

Volume = V1= 7.55L                                                    Volume = V2= 3.81 L

Temperature = T1= 54.0C=54.0+273=327K                 Temp = T2=

application of Ideal gas equation

        P1V1/T1 = P2V2/T2

     0.530 x 7.55/327 = 0.710 x3.81/T2

T2= 221.06 K

Final temp= 221.06K = 221.06 - 273 = - 51.94C

2)

                           Initial conditons                          final conditions

              P1= 360.0torr                                        P2=

             V1= 6.09 L                                              V2= 3.81 L

              T1= 64.0C=64.0+273=337K                  T2= 6C= 6 + 273= 279 K

                   P1V1/T1 = P2V2/T2

                  360.0 x6.09/337 = P2 x3.81/279

                360.0x6.09x279/337x3.81 = P2

P2= 476.397 torr = 476.4 torr

resultant pressure = P2= 476.4 torr

3)

                 Initial coditons                                       Final conditions

         P1 = 757 mm                                                  P2= 501 mm

        V2= 849L                                                         V2=

         T2= 21.9C = 21.9+273= 294.9K                       T2= - 8.1C = - 8.1+273 = 264.9K

              P1V1/T1= P2V2/T2

              757 x849/294.9 = 510 xV2/264.9

V2= 1131.98L