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
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]
