Respuesta :
Answer:
A. [tex]W=600\ J[/tex]
B. [tex]Q=2112\ J[/tex]
C. [tex]\Delta U=1512\ J[/tex]
D. [tex]W=0\ J[/tex]
Explanation:
Given:
- no. of moles of oxygen in the cylinder, [tex]n=0.2[/tex]
- initial pressure in the cylinder, [tex]P_i=2\times 10^5\ Pa[/tex]
- initial temperature of the gas in the cylinder, [tex]T_i=360\ K[/tex]
According to the question the final volume becomes twice of the initial volume.
Using ideal gas law:
[tex]P.V=n.R.T[/tex]
[tex]2\times 10^5\times V_i=0.2\times 8.314\times 360[/tex]
[tex]V_i=0.003\ m^3[/tex]
A.
Work done by the gas during the initial isobaric expansion:
[tex]W=P.dV[/tex]
[tex]W=P_i\times (V_f-V_i)[/tex]
[tex]W=2\times 10^5\times (0.006-0.003)[/tex]
[tex]W=600\ J[/tex]
C.
we have the specific heat capacity of oxygen at constant pressure as:
[tex]c_v=21\ J.mol^{-1}.K^{-1}[/tex]
Now we apply Charles Law:
[tex]\frac{V_i}{T_i} =\frac{V_f}{T_f}[/tex]
[tex]\frac{0.003}{360} =\frac{0.006}{T_f}[/tex]
[tex]T_f=720\ K[/tex]
Now change in internal energy:
[tex]\Delta U=n.c_p.(T_f-T_i)[/tex]
[tex]\Delta U=0.2\times 21\times (720-360)[/tex]
[tex]\Delta U=1512\ J[/tex]
B.
Now heat added to the system:
[tex]Q=W+\Delta U[/tex]
[tex]Q=600+1512[/tex]
[tex]Q=2112\ J[/tex]
D.
Since during final cooling the process is isochoric (i.e. the volume does not changes). So,
[tex]W=0\ J[/tex]
