Answer:
(d) 2996 kJ
Explanation:
We have given initial volume [tex]v_1=2m^3[/tex]
initial pressure [tex]p_1=500\ kPa[/tex]
initial temperature [tex]T_1=300\ K[/tex]
we know that during isothermal process the work done is given by
[tex]W=p_1v_1ln\frac{v_2}{v_1}[/tex]
where [tex]v_2= final\ volume =0.1m^3\ given[/tex]
putting all these value in formula of work done
[tex]W=500\times 2\times ln\frac{0.1}{2}[/tex]
=-2995.8 kJ here negative sign indicates that work is dine on the gas
so wok done =2995.8 kJ