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Oduvanchick [21]
2 years ago
5

I need help ASAP!

Physics
1 answer:
san4es73 [151]2 years ago
7 0

Answer:

b.

Explanation:

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3 years ago
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
ikadub [295]

Answer:

A. W=600\ J

B. Q=2112\ J

C. \Delta U=1512\ J

D. W=0\ J

Explanation:

Given:

  • no. of moles of oxygen in the cylinder, n=0.2
  • initial pressure in the cylinder, P_i=2\times 10^5\ Pa
  • initial temperature of the gas in the cylinder, T_i=360\ K

<em>According to the question the final volume becomes twice of the initial volume.</em>

<u>Using ideal gas law:</u>

P.V=n.R.T

2\times 10^5\times V_i=0.2\times 8.314\times 360

V_i=0.003\ m^3

A.

<u>Work done by the gas during the initial isobaric expansion:</u>

W=P.dV

W=P_i\times (V_f-V_i)

W=2\times 10^5\times (0.006-0.003)

W=600\ J

C.

<u>we have the specific heat capacity of oxygen at constant pressure as:</u>

c_v=21\ J.mol^{-1}.K^{-1}

Now we apply Charles Law:

\frac{V_i}{T_i} =\frac{V_f}{T_f}

\frac{0.003}{360} =\frac{0.006}{T_f}

T_f=720\ K

<u>Now change in internal energy:</u>

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

\Delta U=0.2\times 21\times (720-360)

\Delta U=1512\ J

B.

<u>Now heat added to the system:</u>

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

7 0
3 years ago
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