A cylinder with a piston contains 0.200 mol of nitrogen at 1.50×105 Pa and 320 K . The nitrogen may be treated as an ideal gas. The gas is first compressed isobarically to half its original volume. It then expands adiabatically back to its original volume, and finally it is heated isochorically to its original pressure. Find the heat added to the gas during the final heating. Find the internal-energy change of the gas during the final heating.

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aristocles aristocles · Answer 1 · 2019-10-02T08:58:03+02:00

Answer:

[tex]Q = -105 J[/tex]

Also we know that for cyclic process change in internal energy is always ZERO

Explanation:

First gas is compressed isobarically such that its volume is half of initial volume

So its temperature is also half

So heat given in this process is given as

[tex]Q = nC_p \Delta T[/tex]

for diatomic gas we have

[tex]C_p = \frac{7}{2} R[/tex]

so we will have

[tex]Q = 0.200(\frac{7}{2}R)(160 - 320)[/tex]

[tex]Q = -930.7 J[/tex]

Now in adiabatic process heat is not transferred

so in this process

[tex]Q = 0[/tex]

so we have

[tex]T_1V_1^{1.4-1} = T_2V_2^{1.4-1}[/tex]

[tex](160)(\frac{V}{2})^{0.4} = T_2(V)^{0.4}[/tex]

[tex]T_2 = 121.26 K[/tex]

Now it is again reached to original pressure

so temperature will become initial temperature

so heat given in that part

[tex]Q_3 = nC_v\Delta T[/tex]

here we know that

[tex]C_v = \frac{5}{2}R[/tex]

[tex]Q_3 = (0.200)(\frac{5}{2}R)(320 - 121.26)[/tex]

[tex]Q_3 = 825.76 J[/tex]

So total heat given to the system is

[tex]Q = -930.7 + 0 + 825.76[/tex]

[tex]Q = -105 J[/tex]

Also we know that for cyclic process change in internal energy is always ZERO