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An ideal monatomic gas expands isobarically from state A to state B. It is then compressed isothermally from state B to state C and finally cooled at constant volume until it returns to its initial state A.
[tex]V_A[/tex] = 4 x 10⁻³[tex]V_B[/tex] = 8 x 10⁻³[tex]P_A[/tex] = [tex]P_B[/tex] = 1 x 10⁶ [tex]P_C[/tex] = 2 x 10⁶[tex]T_A[/tex] = 600 k
What is the temperature of the gas when it is in state B?
How much work is done by the gas in expanding isobarically from A to B?
How much work is done on the gas in going from B to C?
The graph is a PV diagram. It does not show any other numbers on it other than what is given.

Respuesta :

skyluke89

1) The temperature of the gas in state B is 1200 K

2) The work done by the gas from A to B is 4000 J

3) The work done by the gas from B to C is -5546 J

Explanation:

1)

The temperature of the gas when it is in state B can be found by using the ideal gas equation:

[tex]\frac{P_A V_A}{T_A}=\frac{P_B V_B}{T_B}[/tex]

where

[tex]P_A = 1\cdot 10^6 Pa[/tex] is the pressure in state A

[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A

[tex]T_A = 600 K[/tex] is the temperature in state A

[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

[tex]T_B[/tex] is the temperature in state B

Solving for [tex]T_B[/tex],

[tex]T_B=\frac{P_B V_B T_A}{P_A V_A}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})(600)}{(1\cdot 10^6)(4\cdot 10^{-3})}=1200 K[/tex]

2)

The work done by a gas during an isobaric transformation (as the one between A and B) is

[tex]W=p\Delta V = p (V_B-V_A)[/tex]

where

p is the pressure (which is constant)

[tex]V_B[/tex] is the final volume

[tex]V_A[/tex] is the initial volume

Here we have:

[tex]p=1\cdot 10^6 Pa[/tex]

[tex]V_A = 4 \cdot 10^{-3} m^3[/tex] is the volume in state A

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

Substituting,

[tex]W=(1\cdot 10^6)(8\cdot 10^{-3}-4\cdot 10^{-3})=4000 J[/tex]

3)

During an isothermal expansion, the produce between the pressure of the gas and its volume remains constant (Boyle's law), so we can write:

[tex]P_BV_B = P_C V_C[/tex]

where

[tex]P_B = 1\cdot 10^6 Pa[/tex] is the pressure in state B

[tex]V_B = 8\cdot 10^{-3} m^3[/tex] is the volume in state B

[tex]P_C = 2\cdot 10^6 Pa[/tex] is the pressure in state C

[tex]V_C[/tex] is the volume in state C

Solving for [tex]V_C[/tex],

[tex]V_C = \frac{P_B V_B}{P_C}=\frac{(1\cdot 10^6)(8\cdot 10^{-3})}{2\cdot 10^6}=4\cdot 10^{-3} m^3[/tex]

The work done by a gas during an isothermal transformation is given by

[tex]W=nRT ln(\frac{V_C}{V_B})[/tex] (1)

where

n is the number of moles

[tex]R=8.314 J/mol K[/tex] is the gas constant

T is the constant temperature (in this case, [tex]T_B[/tex])

[tex]V_C, V_B[/tex] are the final and initial volume, respectively

The number of moles of the gas can be found as

[tex]n=\frac{p_A V_A}{RT_A}=\frac{(1\cdot 10^6)(4\cdot 10^{-3})}{(8.314)(600)}=0.802 mol[/tex]

So now we can use eq.(1) to find the work done by the gas from B to C:

[tex]W=(0.802)(8.314)(1200) ln(\frac{4\cdot 10^{-3}}{8\cdot 10^{-3}})=-5546 J[/tex]

And the work is negative because the gas has contracted, so the work has been done by the surrounding on the gas.

Learn more about ideal gases:

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