Question

Air confined to one side of a rigid container divided by a partition, as shown in Fig.2.The outer side is initially evacuated.The air is initially at p₁=5 bar,T₁=500 K, and V₁= 0.2m³. When the partition is removed, the air expands to fill the entire chamber. Measurements show that V₂=2V₁and p₂=p₁/4.Assuming the air behaves as an ideal gas, determine (a) the final temperature, in K, and (b)the heat transfer, kJ.

Answer 1

Draw the rigid container system as shown below:    

(a)

Calculate the temperature at state point 2 by using the ideal gas equation

$$ p_{1} V_{1}=m R T_{1} $$

\(p_{2} V_{2}=m R T_{2}\)

\(\frac{p_{1} V_{1}}{T_{1}}=\frac{p_{2} V_{2}}{T_{2}}\)

Here, \(p_{1}\) is the pressure at state point \(1, V_{1}\) is the volume at state point \(1, T_{1}\) is the temperature at sate point \(1, p_{2}\) is the pressure at state point \(2, V_{2}\) is the volume at state point \(2, T_{2}\) is the temperature at sate point 2 .

Substitute \(0.2 \mathrm{~m}^{3}\) for \(V_{1}, 5\) bar for \(p_{1}, 500 \mathrm{~K}\) for \(T_{1}, \frac{1}{4} p_{1}\) for \(p_{2}\) and \(2 V_{1}\) for \(V_{2}\)

\(\frac{5 \times 0.2}{500}=\frac{\frac{1}{4}(5) 2(0.2)}{T_{2}}\)

\(T_{2}=250 \mathrm{~K}\)

Therefore, temperature at state point 2 is \(\overline{250 \mathrm{~K}}\)

(b)

From properties of air table,

At \(T_{1}=500 \mathrm{~K}\)

Internal energy at sate point1 is \(u_{1}=359.49 \mathrm{~kJ} / \mathrm{kg}\)

And at \(T_{2}=250 \mathrm{~K}\)

Internal energy at state point \(2 u_{2}=178.28 \mathrm{~kJ} / \mathrm{kg}\)

Calculate the Mass of air,

\(m=\frac{p_{1} V_{1}}{R T_{1}}\)

Here, \(R\) is the gas constant of the air

Substitute \(0.287 \mathrm{~kJ} / \mathrm{kg} \cdot \mathrm{K}\) for \(R\)

\(\begin{aligned} m &=\frac{p_{1} V_{1}}{R T_{1}} \\ &=\frac{5 \times 10^{2} \times 0.2}{0.287 \times 500} \\ &=0.6968 \mathrm{~kg} \end{aligned}\)

Calculate the rate of heat transfer,

$$ \begin{aligned} Q &=W+\Delta U \\ &=0+m\left(u_{2}-u_{1}\right) \end{aligned} \quad(\because \text { there is no moving boundary }) $$

$$ =m\left(u_{2}-u_{1}\right) $$

Substitute \(0.6968 \mathrm{~kg}\) for \(m, 178.28 \mathrm{~kJ} / \mathrm{kg}\) for \(u_{2}\)

$$ \begin{aligned} Q &=0.6968(178.28-359.49) \\ &=-126.267 \mathrm{~kJ} \end{aligned} $$

Therefore, rate of heat transfer is -126.267 kJ.

Answer 2

Is it a lump or a bump?