Q#19.40
One-half mole of an ideal gas is taken from state a to state c, as shown in Fig. P19.40.
(a) Calculate the final temperature of the gas.
(b) Calculate the work done on (or by) the gas as it moves from state a to state c.
(c) Does heat leave the system or enter the system during this process? How much heat? Explain.
Answer:
Use pV = nRT to calculate T. W is the area under the process in the pV-diagram. Use ΔU = n$C_v\Delta T$ and ΔU =Q − W to calculate Q.
Given: In state c: $p_c=2.0 \times 10^5 \ Pa$ and $V_c=0.0040 \ m^3$
In state a: $p_a=4.0 \times 10^5 \ Pa$ and $V_c=0.0020 \ m^3$
(a) $T_c=\frac{P_cV_c}{nR}$
$T_c=\frac{(2.0 \times 10^5 \ Pa)(0.0040 \ m^3)}{(0.0500 \ mola)(8.3145 \ J/mol.K)}$ = 192 K
(b) $W_{ac}=W_{ab}+W_{bc}$
$W_{ab}=\frac{1}{2}(4.0 \times 10^5 \ Pa+2.0 \times 10^5 \ Pa)(0.0030 \ m^3-0.0020 \ m^3)=300 J$
$W_{ac}=(2.0 \times 10^5 \ Pa)(0.0040 \ m^3-0.0030 \ m^3)=200 J$, so
$W_{ac}$=300 J + 200 J = +500 J
500 J of work is done by the gas.
(c) $T_a=\frac{P_aV_a}{nR}$
$T_a=\frac{(4.0 \times 10^5 \ Pa)(0.0020 \ m^3)}{(0.0500 \ mola)(8.3145 \ J/mol.K)}$ = 192 K
For the process, ΔT = 0, so ΔU = 0 and Q = W = +500 J.
500 J of heat enters the system.
The work done by the gas is positive since the volume increases.
Q#19.41
When a system is taken from state a to state b in Fig. P19.41 along the path acb, 90.0 J of heat flows into the system and 60.0 J of work is done by the system.
(a) How much heat flows into the system along path adb if the work done by the system is 15.0 J?
(b) When the system is returned from b to a along the curved path, the absolute value of the work done by the system is 35.0 J. Does the system absorb or liberate heat? How much heat?
(c) If $U_a$ = 0 and $U_d$ = 8.0 J, find the heat absorbed in the processes ad and db.
Answer:
Use ΔU = Q − W and the fact that ΔU is path independent. W > 0 when the volume increases, W < 0 when the volume decreases, and W = 0 when the volume is constant. Q > 0 if heat flows into the system.
The paths are sketched in Figure 19.41.
$Q_{acb}$ = +90.0 J (positive since heat flows in) 60 0 J, $W_{acb}$ = +60.0 J. (positive since ΔV > 0)
(a) ΔU = Q − W
ΔU is path independent; Q and W depend on the path.
ΔU= $U_b-U_a$ This can be calculated for any path from a to b, in particular for path acb:
$\Delta U_{a→b}=Q_{acb}-Q_{acb}=90.0 \ J - 60.0 \ J =30.0 \ J$
Now apply ΔU = Q − W to path adb; ΔU = 30.0 J; ΔU for this path also. 15 0 J
$W_{adb}$ = +15.0 J (positive since ΔV > 0)
$\Delta U_{a→b}=Q_{adb}-W_{adb}$, so
$Q_{adb} = \Delta U_{a→b}+W_{adb}$
$Q_{adb}$ = 30.0 J + 15.0 J = 45.0 J
(b) Apply ΔU = Q − W to path ba: $\Delta U_{b→a}=Q_{ba}-W_{ba}$
$W_{ba}=-35 \ J$ (negative since ΔV < 0)
$\Delta U_{b→a}=U_{a}-U_{b}$
($Q_{ba}$ < 0 the system liberates heat.)
(c) $U_a$ = 0; $U_d$ = 8.0 J
$\Delta U_{a→b}=U_{b}-U_{a}$, so, $U_b=+30 \ J$
process a → d
$U_{a→d}=Q_{ad}-W_{ad}$
$U_{a→d}=U_d-U_a= +8.0 \ J$
$W_{adb}$ = +15.0 J and $W_{adb}=W_{ad}+W_{db}$.
But the work $W_{db}$ for the process d → b is zero since ΔV = 0 for that process. Therefore $W_{ad}=W_{adb}$ = 15.0 J
Then $Q_{ad}=\Delta U_{a→d}+W_{ad}$
$Q_{ad}=+8.0 \ J+15.0 \ J=+23.0 \ J$ (positive implies heat absorbed). process d → b.
$U_{d→b}=Q_{db}-W_{db}$
$W_{db}$ = 0, as already noted.
$U_{d→b}=U_b-U_d= 30.0 \ J -8.0 \ J= +22.0 \ J$
$Q_{db}=\Delta U_{d→b}+W_{db}=+22.0 \ J$ (positive; heat absorbed).
The signs of our calculated $Q_{ad}$ and $Q_{db}$ agree with the problem statement that heat is absorbed in these processes.
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