Known Two tanks containing different amounts of carbon monoxide gas at initially different states are connected by a valve. The valve is opened and the gas allowed to mix while receiving energy by heat transfer. The final equilibrium temperature is known.
Find Determine the final pressure and the heat transfer for the
process.
Schematic and Given Data:
Engineering Model
1. The total amount of carbon monoxide gas is a closed system.
1 2. The gas is modeled as an ideal gas with constant c_{v}.
3. The gas initially in each tank is in equilibrium. The final state is an equilibrium state.
4. No energy is transferred to, or from, the gas by work.
5. There is no change in kinetic or potential energy.
Analysis
a. The final equilibrium pressure p_{ f } can be determined from the ideal gas equation of state
p_{ f }=\frac{m R T_{ f }}{V}
where m is the sum of the initial amounts of mass present in the two tanks, V is the total volume of the two tanks, and T_{ f } is the final equilibrium temperature. Thus,
p_{ f }=\frac{\left(m_{1}+m_{2}\right) R T_{ f }}{V_{1}+V_{2}}
Denoting the initial temperature and pressure in tank 1 as T_{1} \text { and } p_{1}, respectively, V_{1}=m_{1} R T_{1} / p_{1}. Similarly, if the initial temperature and pressure in tank 2 are T_{2} \text { and } p_{2}, V_{2}=m_{2} R T_{2} / p_{2}. Thus, the final pressure is
p_{ f }=\frac{\left(m_{1}+m_{2}\right) R T_{ f }}{\left(\frac{m_{1} R T_{1}}{p_{1}}\right)+\left(\frac{m_{2} R T_{2}}{p_{2}}\right)}=\frac{\left(m_{1}+m_{2}\right) T_{ f }}{\left(\frac{m_{1} T_{1}}{p_{1}}\right)+\left(\frac{m_{2} T_{2}}{p_{2}}\right)}
Inserting values
p_{ f }=\frac{(10 kg )(315 K )}{\frac{(2 kg )(350 K )}{0.7 bar }+\frac{(8 kg )(300 K )}{1.2 bar }}=1.05 bar
b. The heat transfer can be found from an energy balance, which reduces with assumptions 4 and 5 to give
\Delta U=Q-W^{\nearrow0}
or
Q=U_{ f }-U_{ i }
U_{ i } is the initial internal energy, given by
U_{ i }=m_{1} u\left(T_{1}\right)+m_{2} u\left(T_{2}\right)
where T_{1} \text { and } T_{2} are the initial temperatures of the CO in tanks 1 and 2, respectively. The final internal energy is U_{ f }
U_{ f }=\left(m_{1}+m_{2}\right) u\left(T_{ f }\right)
Introducing these expressions for internal energy, the energy balance becomes
Q=m_{1}\left[u\left(T_{ f }\right)-u\left(T_{1}\right)\right]+m_{2}\left[u\left(T_{ f }\right)-u\left(T_{2}\right)\right]
Since the specific heat c_{v} is constant (assumption 2)
Q=m_{1} c_{v}\left(T_{ f }-T_{1}\right)+m_{2} c_{v}\left(T_{ f }-T_{2}\right)
Evaluating c_{v} as the average of the values listed in Table A-20 at 300 K and 350 K, c_{v}=0.745 kJ / kg \cdot K. Hence,
Q=(2 kg )\left(0.745 \frac{ kJ }{ kg \cdot K }\right)(315 K -350 K )
+(8 kg )\left(0.745 \frac{ kJ }{ kg \cdot K }\right)(315 K -300 K )
=+37.25 kJ
The plus sign indicates that the heat transfer is into the system.
1 By referring to a generalized compressibility chart, it can be verified that the ideal gas equation of state is appropriate for CO in this range of temperature and pressure. Since the specific heat c_{v} of CO varies little over the temperature interval from 300 to 350 K (Table A-20), it can be treated as constant with acceptable accuracy.
Skills Developed
Ability to…
• define a closed system and identify interactions on its boundary.
• apply the energy balance using the ideal gas model when the specific heat c_{v} is constant.
Quick Quiz
Evaluate Q using specific internal energy values for CO from Table A-23. Compare with the result using constant c_{v}. Ans. 36.99 kJ
