Question #089c7

1 Answer
May 10, 2017

Change in internal energy, #Delta# U = #-143.56J#

Explanation:

Using the first law of thermodynamics
#Delta# U → #Delta# Q + #Delta# W
where #Delta# U is the change in internal energy of a system
#Delta# Q is the energy supplied to the system
#Delta# W is the work done on or by the system

#Delta# W = p#Delta#V
#Delta# W= 101.325 Pa (5.6#xx10^-3m^3- 1.21xx10^-3m^3#)
#Delta# W= 0.44 J

Since work is being done by the gas #Delta# W is negative
i.e. #Delta# W= -0.44 J

Since energy is being released by the gas, #Delta#Q is negative
i.e. #Delta# Q= -124 J

Change in internal energy, #Delta# U= #-124 + (-0.44) = -143.56J#
So internal energy is decreasing.

*** I converted the L to #m^3# and the atm to Pa