Question

A cylinder with a piston contains 0.200 mol of oxygen at 2.00×105 Pa and 360 K . The oxygen may be treated as an ideal gas. The gas first expands isobarically to twice its original volume. It is then compressed isothermally back to its original volume, and finally it is cooled isochorically to its original pressure.Part AFind the work done by the gas during the initial expansion.W initial = JPart BFind the heat added to the gas during the initial expansion.Q initial = JPart CFind internal-energy change of the gas during the initial expansion.ΔU initial = JPart DFind the work done during the final cooling;W final = J

Answer 1

A.
`W=600\ J`

B.
`Q=2112\ J`

C.
`\Delta U=1512\ J`

D.
`W=0\ J`

Step-by-step explanation:

Given:

• no. of moles of oxygen in the cylinder,
  `n=0.2`

• initial pressure in the cylinder,
  `P_i=2* 10^5\ Pa`

• initial temperature of the gas in the cylinder,
  `T_i=360\ K`

According to the question the final volume becomes twice of the initial volume.

Using ideal gas law:

`P.V=n.R.T`

`2* 10^5* V_i=0.2* 8.314* 360`

`V_i=0.003\ m^3`

A.

Work done by the gas during the initial isobaric expansion:

`W=P.dV`

`W=P_i* (V_f-V_i)`

`W=2* 10^5* (0.006-0.003)`

`W=600\ J`

C.

we have the specific heat capacity of oxygen at constant pressure as:

`c_v=21\ J.mol^(-1).K^(-1)`

Now we apply Charles Law:

`(V_i)/(T_i) =(V_f)/(T_f)`

`(0.003)/(360) =(0.006)/(T_f)`

`T_f=720\ K`

Now change in internal energy:

`\Delta U=n.c_p.(T_f-T_i)`

`\Delta U=0.2* 21* (720-360)`

`\Delta U=1512\ J`

B.

Now heat added to the system:

`Q=W+\Delta U`

`Q=600+1512`

`Q=2112\ J`

D.

Since during final cooling the process is isochoric (i.e. the volume does not changes). So,

`W=0\ J`