Question

Question #089c7

Answer 1

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