# A monatomic ideal gas expands slowly to twice its original

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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SlabydouluS62

Step 1
Solution
a) For the isothermal process the temperature is constant during the process which means the change in temperature $\mathrm{△}T=0$. As the change in internal energy depends directly of $\mathrm{△}T$, we could obtain $\mathrm{△}U$ equals zero by
$\mathrm{△}U=n{C}_{v}\mathrm{△}T=0$
Now plug the values for W and $\mathrm{△}U$ into equation (1) to get Q when the process is isothermal.
$Q=W+\mathrm{△}U$
$=450j+0$
$=450j$
For the isothermal process all heat goes to increase the work done.

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 $\mathrm{△}U$ by plug our values for W and Q into equation (1) to get $\mathrm{△}U$ for the adiabatic process where $Q=0$
$\mathrm{△}U=Q-W$
$=0-450j$
$=-450j$
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
$Q=n{C}_{p}\mathrm{△}T$ (2)
Where n is the number of moles and ${C}_{p}$ is the molar heat capacity at constant pressure and for the monatomic gas ${C}_{p}=\frac{5}{2}R$ where R is the gas constant. To calculate the heat Q we want to exclude the term $n\mathrm{△}T$ because we do not have their values. At constant pressure the work done is given by
$W=p\mathrm{△}V$ (using ideal gas equation)
$W=nR\mathrm{△}T$ (solve for $\mathrm{△}T$)
$\mathrm{△}T=\frac{W}{nR}$
Now use the value of $\mathrm{△}T$ in equation (2) and plug the values for W and ${C}_{p}$ to get the heat Q
$Q=n{C}_{p}\mathrm{△}T$
$=\text{⧸}n\left(\frac{}{}\text{⧸}R\right)\frac{W}{\text{⧸}nR}$
$=\frac{5}{2}×450J$
$=1125J$
Now we can use these values for Q and W and plug it into equation (1) to get $\mathrm{△}U$ for the isobaric process
$\mathrm{△}U=Q-W$
$=1125J-450J$
$=675J$
The heat added in the isobaric process goes to increase the work done by the gas and its internal energy.

Maricela Alarcon

a)
b)
Explanation:
To find the heat transferred in the process let's use the first law of thermodynamics
$\mathrm{△}E=Q+W$
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 $\left(Q=0\right)$
$\mathrm{△}E=W$
$\mathrm{△}E=450J$
$Q=0$
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
$W=-P\mathrm{△}V$
$P=W\mathrm{△}V$
$P=\frac{450}{2V-V}$
$P=\frac{450}{V}$
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
$PV1=mRT1$
$PV2=nRT2$
$P\left(v2-V1\right)=nR\left(T2-T1\right)$
$450=nR\left(T2-T1\right)$
$\mathrm{△}T=\frac{450}{n}R$
We can use the Boltzmann equation that relates energy and temperature
$E=KT$
$\mathrm{△}E=K\mathrm{△}T$
$\mathrm{△}E=K\frac{450}{n}R$
Boltzmann's constant is
$K=\frac{R}{n}$
replacing
$\mathrm{△}E=\left(\frac{R}{n}\right)\frac{450}{n}R$
$\mathrm{△}E=450{n}^{2}$
We already have the two terms of the first law of thermodynamics
$Q=DE-W$
$Q=\frac{450}{{n}^{2}}-450$
$Q=450\left(\frac{1}{{n}^{2}}-1\right)$

nick1337

Part (a)
In this case, for the isothermal process, the temperature is constant.
The change in temperature,
$\mathrm{△}T=0$
The change in energy of the gas,
$\mathrm{△}U=n{C}_{v}\mathrm{△}T$
$=0J$
Step 2
NSK
The heat added to the gas is,
$Q=W+\mathrm{△}U$
$=450J+0J$
$=450J$
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,
$\mathrm{△}U=Q-W$
$0J-450J$
$=-450J$
Part (c)
In this сase, for the isobaric process, the pressure is constant.
When the pressure is constant, the heat added is
$Q=n{C}_{p}\mathrm{△}T$
$=n\left(\frac{5}{2}R\right)\left(\frac{W}{nR}\right)$
$=\frac{5}{2}\left(450J\right)$
$=1125J$
Step 5
The change in energy of the gas,
$\mathrm{△}U=Q-W$
$=1125J-450J$
$=675J$