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1. A sample of ideal gas at room temperature occupies a volume of 38.0 L at a pr

ID: 500839 • Letter: 1

Question

1. A sample of ideal gas at room temperature occupies a volume of 38.0 L at a pressure of 512 torr . If the pressure changes to 2560 torr , with no change in the temperature or moles of gas, what is the new volume, V2?

Express your answer with the appropriate units.

V2 =

2. If the volume of the original sample in Part A (P1 = 512 torr , V1 = 38.0 L ) changes to 51.0 L , without a change in the temperature or moles of gas molecules, what is the new pressure, P2?

Express your answer with the appropriate units.

P2 =

3. A sample of gas in a balloon has an initial temperature of 8.0 C and a volume of 1.38×103 L . If the temperature changes to 73 C , and there is no change of pressure or amount of gas, what is the new volume, V2, of the gas?

Express your answer with the appropriate units.

V2 =

4. What Celsius temperature, T2, is required to change the volume of the gas sample in Part A (T1 = 8.0 C , V1= 1.38×103 L ) to a volume of 2.76×103 L ? Assume no change in pressure or the amount of gas in the balloon.

Express your answer with the appropriate units.

T2 =

5. If the initial volume of the balloon is 2.5 L , what will its volume be after it cools? (Assume constant pressure.)

Express your answer using two significant figures.

V2 =

Explanation / Answer

1.

Initial                                                                      Final

P1 = 512 torr                                                        P2 = 2560 torr

V1    = 38L                                                           V2 =

    P1V1   = P2V2

     V2       = P1V1/P2

                = 512*38/2560 = 7.6L

2.

P1 = 512 torr                                                        P2 = torr

V1    = 38L                                                           V2 = 51L

    P1V1   = P2V2

     P2     = P1V1/V2

              = 512*38/51 = 381.5torr

3.

Initial                                                      Final

V1    = 1.38*10^3L                                   V2 =

T1    = 8C0 = 8+273 = 281K                   T2   = 73C0 = 73 + 273 = 346K

   V1/T1     =   V2/T2

   V2           = V1T2/T1

                 = 1380*346/281   = 1699.2L

4.

Initial                                                      Final

V1    = 1.38*10^3L                                   V2 =2.76*10^3  

T1    = 8C0 = 8+273 = 281K                   T2   =

   V1/T1     =   V2/T2

    T2           = V2T1/V1

                   = 2760*281/1380 = 562K