I'm studying for an exam in thermodynamics and therefore going through old exams, but I am stuck on one assignment.
One kilomole of ideal monoatomic gas under standard conditions is
processed from condition 1 to condition 2 in two different ways: $1 \rightarrow 3 \rightarrow 2$ and $1 \rightarrow 4 \rightarrow 2$. Find the ratio of the thermal energies that must be transferred to the gas in these processes.
I tried to solve this by first using the first law of thermodynamics
$dU=dQ+dW$
where U is the internal energy, Q the thermal energy and W the work done by the gas.
The work can be obtained from
$dW=-pdV$
which in this case means the area under the graph
For the different pathways we get the following work:
$1 \rightarrow 3, \Delta W=0$
$3 \rightarrow 2, \Delta W=2p_0V_0$
$1 \rightarrow 4, \Delta W=p_0V_0$
$4 \rightarrow 2, \Delta W=0$
Since the gas is monoatomic and ideal $U$ can be calculated from
$U=3Nk_BT/2=3p_0V_0/2$
Plugging this into the first law of thermodynamics would give me
$1 \rightarrow 3 \rightarrow 2: Q=1.5p_0V_0-2p_0V_0=-1,5p_0V_0$
$1 \rightarrow 4 \rightarrow 2: Q=1.5 p_0V_0-p_0V_0=0.5p_0V_0$
However the correct answer is
$1 \rightarrow 3 \rightarrow 2: Q=1.5p_0V_0+5p_0V_0=6.5p_0V_0$
$1 \rightarrow 4 \rightarrow 2: Q=2.5p_0V_0+3p_0V_0=5.5p_0V_0$
What have I done wrong? It seems like I have missed some terms in $U$ but where would they come from?
Best Answer
The equation for First Law of Thermodynamics can be written in the form (starting from the equation you wrote with $dW=-pdV$)
$$dQ=dU+pdV$$ where $dU=C_v dT$.Here for mono atomic ideal gas $C_v=\frac{3}{2}K_BT$. Using equation of state for ideal gas one gets $$ dU=\frac{3}{2}[Vdp+pdV]$$ So $$1\rightarrow3:dU=\frac{3}{2}p_0V_0,pdV=0$$ $$3\rightarrow2:dU=3p_0V_0,pdV=2p_0V_0$$ $$1\rightarrow4:dU=\frac{3}{2}p_0V_0,pdV=p_0V_0$$ $$4\rightarrow2:dU=3p_0V_0,pdV=0$$ So we see that the thermal energy for the two different path is $$1\rightarrow3\rightarrow 2:dQ=(\frac{3}{2}+3+2)p_0V_0=6.5p_0V_0$$ $$1\rightarrow4\rightarrow 2:dQ=(1+\frac{3}{2}+3)p_0V_0=5.5p_0V_0$$ Note:Internal energy is a state function which means it does not depends on which path you choose.You can see here in both path $dU$ is equal.