A monatomic ideal gas expands slowly to twice its original

Pam Stokes 2021-12-17 Answered
A monatomic ideal gas expands slowly to twice its original volume, doing 450 J of work in the process. Find the heat added to the gas and the change in internal energy of the gas if the process is (a) isothermal; (b) adiabatic; (c) isobaric.
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Expert Answer

SlabydouluS62
Answered 2021-12-18 Author has 52 answers

Step 1
Solution
a) For the isothermal process the temperature is constant during the process which means the change in temperature T=0. As the change in internal energy depends directly of T, we could obtain U equals zero by
U=nCvT=0
Now plug the values for W and U into equation (1) to get Q when the process is isothermal.
Q=W+U
=450j+0
=450j
For the isothermal process all heat goes to increase the work done.
Q=450j and U=0
Step 2
b) For the adiabatic process, the gas is insulated and no heat added to the system so heat Q is equal zero
Q=0
Now we can find the change in internal energy U by plug our values for W and Q into equation (1) to get U for the adiabatic process where Q=0
U=QW
=0450j
=450j
In the adiabatic process, the added heat is zero, and the energy to do work comes from the internal energy that decreases.
Step 3
c) For the isobaric process the pressure is constant and when the pressure is constant the heat is given by
Q=nCpT (2)
Where n is the number of moles and Cp is the molar heat capacity at constant pressure and for the monatomic gas Cp=52R where R is the gas constant. To calculate the heat Q we want to exclude the term nT because we do not have their values. At constant pressure the work done is given by
W=pV (using ideal gas equation)
W=nRT (solve for T)
T=WnR
Now use the value of T in equation (2) and plug the values for W and Cp to get the heat Q
Q=nCpT
=n(R)WnR
=52×450J
=1125J
Now we can use these values for Q and W and plug it into equation (1) to get U for the isobaric process
U=QW
=1125J450J
=675J
The heat added in the isobaric process goes to increase the work done by the gas and its internal energy.

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Maricela Alarcon
Answered 2021-12-19 Author has 28 answers

Answer:
a) Q=450J E=0
b) E=450J Q=0
Explanation:
To find the heat transferred in the process let's use the first law of thermodynamics
E=Q+W
Let's apply this equation every case.
Isothermal process. This is done at a constant temperature so that internal energy change is zero.
0=Q+W
Q=W
Q=450J
Adiabatic process. In this process the system is isolated, so there is no heat transfer (Q=0)
E=W
E=450J
Q=0
Isobaric process. Work is done by changing the volume at constant pressure. In this case the internal energy and heat changes in the process, so we will calculate with the equation
W=PV
P=WV
P=4502VV
P=450V
From here we can find the temperature using the ideal gas equation
PV=nRT
Let's write this equation for the two volumes, assuming the constant pressure
PV1=mRT1
PV2=nRT2
P(v2V1)=nR(T2T1)
450=nR(T2T1)
T=450nR
We can use the Boltzmann equation that relates energy and temperature
E=KT
E=KT
E=K450nR
Boltzmann's constant is
K=Rn
replacing
E=(Rn)450nR
E=450n2
We already have the two terms of the first law of thermodynamics
Q=DEW
Q=450n2450
Q=450(1n21)

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nick1337
Answered 2021-12-27 Author has 575 answers

Part (a)
In this case, for the isothermal process, the temperature is constant.
The change in temperature,
T=0
The change in energy of the gas,
U=nCvT
=0J
Step 2
NSK
The heat added to the gas is,
Q=W+U
=450J+0J
=450J
Part (b)
In this case, for the adiabatic process, the gas is insulated and no heat added to the system.
The heat added to the system is,
Q=0J
The change in energy of the gas,
U=QW
0J450J
=450J
Part (c)
In this сase, for the isobaric process, the pressure is constant.
When the pressure is constant, the heat added is
Q=nCpT
=n(52R)(WnR)
=52(450J)
=1125J
Step 5
The change in energy of the gas,
U=QW
=1125J450J
=675J

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