Answer:
[tex]W=-6.99kJ[/tex]
Explanation:
Hello,
In this case, since the work at constant pressure is computed via:
[tex]W=P\Delta V=P(V_2-V_1)[/tex]
For an initial volume of 99.0 L and a final one of 98.0 L, accounting for a compression process at a pressure of 69.0 atm, we obtain a negative work, meaning that work is done on the gas, according to the first law of thermodynamics:
[tex]W=69.0atm(98.0L-99.0L)\\\\W=-69atm*L[/tex]
However, the work in kJ with three significant figures because the initial data have all three significant figures, is:
[tex]W=-69.0atm*L*\frac{101.325kPa}{1atm}*\frac{1m^3}{1000L} \\\\W=-6.99kJ[/tex]
Best regards.