Agas expands in volume from 26.7 ml to 89.3 ml at constant temperature. calculate the work done (in joules) if the gas expands (a) against a vacuum, (b) against a constant pressure of 1.5 atm, and (c) against a constant pressure of 2.8 atm.

a. 0 J

b. - 9.5J

c. - 17.8 J

Explanation:

part a  (against a vacumm)

Data given:

initial volume V1 = 26.7 mL

Final volume V2 = 89.3 mL

work done w = ?

temperature = constant

Solution:

formula will be used

             w = - PΔV

where

ΔV = V2 -V1

so the above equation will be written as

          w = - P (V2 - V1) . . . . . . (1)

as we are calculating work against vacuum so the pressure will be 0 atm.

So,

P = 0 atm

Put values in above equation 1

              w = - [(0 atm)(89.3 mL - 26.7 mL)

              w = - [(0 atm)(62.6mL)

               w = - (0)J

               w = 0 J

so the work done will be 0

part b (against constant pressure of 1.5)

Data given:

initial volume V1 = 26.7 mL

Convert mL to L

1000 mL = 1 L

26.7 mL = 26.7/1000 = 0.0267 L

Final volume V2 = 89.3 mL

Convert mL to L

1000 mL = 1 L

89.3 mL = 89.3/1000 = 0.0893 L

work done w = ?

Pressure constant = 1.5 atm

Solution:

formula will be used

             w = - PΔV

where

ΔV = V2 -V1

so the above equation will be written as

          w = - P (V2 - V1) . . . . . . (1)

Put values in above equation 1

              w = - [(1.5 atm)(0.0893 L - 0.0267 L)

              w = - [(1.5 atm)(0.0626 L)

               w = - (0.0939 atm.L)

               w = - 0.0939 atm.L

convert atm.L to J

1 atm.L = 101.3 J

0.0939 atm.L = 0.0939 x 101.3 = 9.5 J

so the work done will be - 9.5 J

Part c ( constant pressure of 2.8 atm)

Data given:

initial volume V1 = 26.7 mL

Convert mL to L

1000 mL = 1 L

26.7 mL = 26.7/1000 = 0.0267 L

Final volume V2 = 89.3 mL

Convert mL to L

1000 mL = 1 L

89.3 mL = 89.3/1000 = 0.0893 L

work done w = ?

Pressure constant = 2.8 atm

Solution:

formula will be used

             w = - PΔV

where

ΔV = V2 -V1

so the above equation will be written as

          w = - P (V2 - V1) . . . . . . (1)

Put values in above equation 1

              w = - [(2.8 atm)(0.0893 L - 0.0267 L)

              w = - [(2.8 atm)(0.0626 L)

               w = - (0.1753 atm.L)

               w = - 0.1753 atm.L

convert atm.L to J

1 atm.L = 101.3 J

0.1753 atm.L = 0.1753 x 101.3 = 17.8 J

so the work done will be - 17.8 J

gasoline is a mixture. it contains several different kinds of compounds that can be present in a variety of ratios.