Question

A sample of helium gas has a volume of 5.30 L at a pressure of 895 mmHg and a temperature of 20 ∘C. A.What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 1550 mL and 339 K if the amount of gas is constant? B. What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 2.45 Land 12 ∘C if the amount of gas is constant? C. What is the pressure of the gas in atm when the volume and temperature of the gas sample are changed to 13.5 Land 40 ∘C if the amount of gas is constant?

Answer 1

A.

Initial                                                                  Final

V1 = 5.3L                                                      V2 = 1550ml = 1.55L

P1 = 895/760     = 1.1776atm                          P2 =

T1 = 20+273 = 293K                                       T2   = 339K

            P1V1/T1    =    P2V2/T2

               P2            = P1V1T2/T1V2

                              = 1.1776*5.3*339/293*1.55   = 4.658atm>>>>answer

B.

Initial                                                                  Final

V1 = 5.3L                                                      V2 = 2.45L

P1 = 895/760     = 1.1776atm                          P2 =

T1 = 20+273 = 293K                                       T2   = 12 +273   = 285K

            P1V1/T1    =    P2V2/T2

               P2            = P1V1T2/T1V2

                               = 1.1776*5.3*285/293*2.45   = 2.478atm

C.

Initial                                                                  Final

V1 = 5.3L                                                      V2 = 13.5L

P1 = 895/760     = 1.1776atm                          P2 =

T1 = 20+273 = 293K                                       T2   = 40 +273   = 313K

            P1V1/T1    =    P2V2/T2

               P2            = P1V1T2/T1V2

                                 = 1.1776*5.3*313/293*13.5   = 0.494atm >>>answere