§ME 231 Homework 7

Sep 25, 2025

§4.78

A rigid tank of volume 2 m3 contains air at 0.21 bar, 290 K. A leak develops and air flows in slowly from the surroundings, which are at 1.1 bar, 312 K. After a while, the tank and its surroundings come to equilibrium. What is the final temperature in the tank, in °C, and how much mass has entered the tank, in kg? Neglect kinetic and potential energy effects and assume the air is an ideal gas with constant specific heats evaluated at 300 K.
V=2m3 V=2m^3
p1=0.21bar p_1=0.21bar
T1=290K T_1=290K
p∞=1.1bar p_\infty=1.1bar
T∞=312K T_\infty=312K
From table A-20 for air where T=300KT=300K:
cp=1.005kJkg∗K c_p=1.005\frac{kJ}{kg*K}
cv=0.718kJkg∗K c_v=0.718\frac{kJ}{kg*K}
k=1.400 k=1.400
Since the pressure reaches equilibrium with the environment (but the same cannot be said for the temperature),
p2=p∞=1.1bar p_2=p_\infty=1.1bar
I will be leveraging the equilibrium equation:
Q−W+min(hin+12Vin2+gzin)−mout(hout+12Vout2+gzout)=E2−E1 Q-W+m_{in}(h_{in}+\frac{1}{2}V_{in}^2+gz_{in})-m_{out}(h_{out}+\frac{1}{2}V_{out}^2+gz_{out})=E_2-E_1
A significant portion of the equation is either 00 or negligible:
minhin=E2−E1 m_{in}h_{in}=E_2-E_1
A few substitutes can be made incrementally. The energy of both states is entirely composed of internal energy because everything else is ignored:
minhin=m2u2−m1u1 m_{in}h_{in}=m_2u_2-m_1u_1
Substitutions using specific heats:
mincpT∞=m2u2−m1u1 m_{in}c_pT_\infty=m_2u_2-m_1u_1
mincpT∞=m2cvT2−m1cvT1 m_{in}c_pT_\infty=m_2c_vT_2-m_1c_vT_1
Here are the knowns (underlined) for now with T2T_2 being the target value I wish to acquire:
mincpT∞‾=m2cv‾T2−m1cvT1‾ m_{in}\underline{c_pT_\infty}=m_2\underline{c_v}T_2-m_1\underline{c_vT_1}
Mass of state 11 can be acquired using the ideal gas equation of state. From table A-1 for air:
M=28.97kgkmol M=28.97\frac{kg}{kmol}
Rˉ=8.3145Jmol∗K \bar{R}=8.3145\frac{J}{mol*K}
R=RˉM=8.3145Jmol∗K28.97kgkmol=287Jkg∗K R=\frac{\bar{R}}{M}=\frac{8.3145\frac{J}{mol*K}}{28.97\frac{kg}{kmol}}=287\frac{J}{kg*K}
pV=mRT pV=mRT
m=pVRT m=\frac{pV}{RT}
m1=p1VRT1=0.21bar∗2m3287Jkg∗K∗290K=0.505kg m_1=\frac{p_1V}{RT_1}=\frac{0.21bar * 2m^3}{287\frac{J}{kg*K} * 290K}=0.505kg
By computing m1m_1, the number of unknowns is even lower now:
mincpT∞‾=m2cv‾T2−m1cvT1‾ m_{in}\underline{c_pT_\infty}=m_2\underline{c_v}T_2-\underline{m_1c_vT_1}
minm_{in} can be calculated right away too by leveraging the fact that T2T_2 cancels out when the ideal gas law is substituted into the equilibrium equation:
p2V=m2RT2 p_2V=m_2RT_2
m2=p2VRT2 m_2=\frac{p_2V}{RT_2}
m2=p2VRT2 m_2=\frac{p_2V}{RT_2}
mincpT∞=m2cvT2−m1cvT1 m_{in}c_pT_\infty=m_2c_vT_2-m_1c_vT_1
mincpT∞=p2VRT2∗cvT2−m1cvT1 m_{in}c_pT_\infty=\frac{p_2V}{RT_2}*c_vT_2-m_1c_vT_1
mincpT∞=p2VcvR−m1cvT1 m_{in}c_pT_\infty=\frac{p_2Vc_v}{R}-m_1c_vT_1
Solving for minm_{in} gets a little messy which is exactly why I use LaTeX\LaTeX lol:
min=p2VcvR−m1cvT1cpT∞=1.1bar∗2m3∗0.718kJkg∗K287Jkg∗K−0.505kg∗0.718kJkg∗K∗290K1.005kJkg∗K∗312K=1.42kg m_{in}=\frac{\frac{p_2Vc_v}{R}-m_1c_vT_1}{c_pT_\infty}=\frac{\frac{1.1bar * 2m^3 * 0.718\frac{kJ}{kg*K}}{287\frac{J}{kg*K}}-0.505kg * 0.718\frac{kJ}{kg*K} * 290K}{1.005\frac{kJ}{kg*K} * 312K}=\boxed{1.42kg}
Lastly, minm_{in} and m2m_2 are related where mout=0m_{out}=0 because nothing leaves the container:
min−mout=m2−m1 m_{in}-m_{out}=m_2-m_1
min=m2−m1 m_{in}=m_2-m_1
m2=min+m1=1.42kg+0.505kg=1.93kg m_2=m_{in}+m_1=1.42kg + 0.505kg=1.93kg
Now everything but T2T_2 is known in the equilibrium equation; hence, T2T_2 can be solved for:
mincpT∞=m2cvT2−m1cvT1 m_{in}c_pT_\infty=m_2c_vT_2-m_1c_vT_1
mincpT∞+m1cvT1=m2cvT2 m_{in}c_pT_\infty+m_1c_vT_1=m_2c_vT_2
T2=mincpT∞+m1cvT1m2cv T_2=\frac{m_{in}c_pT_\infty+m_1c_vT_1}{m_2c_v}
T2=1.42kg∗1.005kJkg∗K∗312K+0.505kg∗0.718kJkg∗K∗290K1.93kg∗0.718kJkg∗K=397.2K T_2=\frac{1.42kg * 1.005\frac{kJ}{kg*K} * 312K + 0.505kg * 0.718\frac{kJ}{kg*K} * 290K}{1.93kg * 0.718\frac{kJ}{kg*K}}=\boxed{397.2K}
I supposed it makes sense for it to be warmer ¯\_(ツ)_/¯