Q#19.45
Two moles of an ideal monatomic gas go through the cycle abc. For the complete cycle, 800 J of heat flows out of the gas. Process ab is at constant pressure, and process bc is at constant volume. States a and b have temperatures $T_a$ = 200 K and $T_b$ = 300 K.
(a) Sketch the pV-diagram for the cycle.
(b) What is the work W for the process ca?
Answer:
Use the 1st law to relate $Q_{tot}$ to $W_{tot}$ for the cycle.
Calculate $W_{ab}$ and $W_{bc}$ and use what we know about $W_{tot}$ to deduce $W_{ca}$.
(a) We aren’t told whether the pressure increases or decreases in process bc.
The two possibilities for the cycle are sketched in Figure 19.45.
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| Fig.19.45 |
For a cycle, ΔU = 0, so $Q_{tot}=W_{tot}$.
The net
heat flow for the cycle is out of the gas, so heat $Q_{tot}$ < 0 and $W_{tot}$ < 0.
Sketch I is correct.
(b) $W_{tot}=Q_{tot}=-800 \ J$
$W_{tot}=W_{ab}+W_{bc}+W_{ca}$
$W_{bc}$ = 0 since ΔV = 0.
$W_{ab}$ = pΔV since p is constant.
But since it is an ideal gas, pΔV= nRΔT.
$W_{ab} = nR(T_b-T_a)$
$W_{ab}=(2.00 \ mol)(8.3145 \ J/mol.K)(300 \ K - 200 \ K) = 1660 \ J$
$-800 \ J=1660 \ J +0+W_{ca}$
$W_{ca}=-2460 \ J$
In process ca the volume decreases and the work W is negative.
Q#19.46
Three moles of an ideal gas are taken around the cycle acb shown in Fig. P19.46. For this gas, $C_p$ = 29.1 J/mol.K. Process ac is at constant pressure, process ba is at constant volume,and process cb is adiabatic. The temperatures of the gas in states a, c, and b are $T_a$ = 300 K, $T_c$ = 492 K and $T_b$ = 600 K. Calculate the total work W for the cycle.
Answer:
Apply the appropriate expression for W for each type of process. pV = nRT and $C_p=C_v+R$
Path ac has constant pressure, so
$W_{ac}=nR(T_c-T_a)$
$W_{ac}=(3.00 \ mol )(8.3145 \ J/mol.K)(492 \ K-300 \ K)=4.789 \times 10^3 J$
Path cb is adiabatic (Q = 0), so
$W_{cb}=Q - \Delta U=-\Delta U=-nC_v\Delta T$ and using $C_v=C_p-R$.
$W_{cb}=-n(C_p-R)\Delta T$
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