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

answers: to make sure that all of the broken pieces of tablet were added to the water, to allow for fair comparison of all three trials, with different particle sizes to avoid experimental error