The First Law of Thermodynamics Problems and Solutions 2

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. 

Fig.19.45

In cycle I, the total work is negative and in cycle II the total work is positive. 

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.

$W_{ab}$ = pΔsince 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 $C_v=C_p-R$.

$W_{cb}=-n(C_p-R)\Delta T$

$W_{cb}=-(3.00 \ mol)(29.1 \ J/mol.K-8.3145 \ J/mol.K)(600 \ K-492 \ K)$

$W_{cb}=-6.735 \times 10^3$ J

Path ba has constant volume, so $W_{ba}$ = 0. So the total work done is 

$W=W_{ab}+W_{cb}+W_{ba}$

$W=4.789 \times 10^3 \ J+(-6.735 \times 10^3 \ J)+0 = -1.95 \times 10^3$ J

W > 0 when ΔV > 0, W < 0 when ΔV < 0 and W = 0 when ΔV = 0.


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