Answer:
The change in internal energy is  [tex]\Delta U = 417331 \ J[/tex]
Explanation:
From the question we are told that
  The initial volume is  [tex]V_i = 5.35 \ L[/tex]
    The final volume is  [tex]V_f = 1.23 \ L[/tex]
   The value of the external pressure is  [tex]P = 1.00 \ atm = 1 * 101325 = 101325 \Pa[/tex]
   The energy released by the gas is  [tex]Q= -128 \ J[/tex]
The negative sign show that energy is released from the system
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Generally  the workdone on the gas is mathematically represented as
     [tex]W = P (V_i - V_f)[/tex]
substituting values
     [tex]W = 101325 *(1.23 - 5.35)[/tex]
    [tex]W = - 4.17*10^{5} \ J[/tex]
The change in internal energy is mathematically evaluated as
     [tex]\Delta U = Q - W[/tex]
substituting values
     [tex]\Delta U = -128 - (-4.17*10^{5})[/tex]
    [tex]\Delta U = 417331 \ J[/tex]
