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Air is compressed from 20 psia and 70 \(^{\circ} \mathrm{F}\) to 150 psia in a compressor. The compressor is operated such that the air temperature remains constant. Calculate the change in the specific volume of air as it passes through this compressor.

Short Answer

Expert verified
Answer: The change in specific volume, ∆v, is given by the expression $$\Delta v = V_1\left(\frac{1}{7.5}-1\right)$$, where V1 is the initial specific volume of the air. The exact value of the change in specific volume cannot be determined without knowing the initial specific volume.

Step by step solution

01

Convert given temperatures to absolute temperatures (Rankine)

To use the ideal gas law, we need to work with absolute temperatures. The given temperature, 70°F, needs to be converted to Rankine. The conversion formula is: $$T_R = T_F + 459.67$$ $$T_1 = 70 + 459.67 = 529.67 \,^{\circ} R$$ Since the temperature remains constant, T2 = 70°F = 529.67 °R.
02

Write the ideal gas law and solve for specific volume

The ideal gas law is given by PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the gas constant, and T is temperature. Since the number of moles and the gas constant remain constant in the process, we can rewrite the equation as: $$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$ We need to find the change in specific volume (∆v = V2 - V1). We can start by rearranging the equation to solve for V2: $$V_2 = V_1 \frac{P_1 T_2}{P_2 T_1}$$
03

Substitute the given values and solve for V2

Now we can plug in the given values to find V2: $$V_2 = V_1 \frac{20 \;psia \cdot 529.67 \,^{\circ} R}{150 \;psia \cdot 529.67\,^{\circ} R}$$ The temperatures in numerator and denominator cancel out: $$V_2 = V_1 \frac{20 \;psia}{150 \;psia}$$ Now we can calculate V2: $$ V_2 = \frac{1}{7.5} V_1 $$
04

Calculate the change in specific volume

We can now calculate the change in specific volume, ∆v = V2 - V1: $$\Delta v = V_2 - V_1$$ Substitute the expression of V2 in terms of V1: $$\Delta v = \frac{1}{7.5}V_1 - V_1$$ Factor out V1: $$\Delta v = V_1\left(\frac{1}{7.5}-1\right)$$ We can't determine the exact value of the change in specific volume since we don't know the initial specific volume (V1). However, we have expressed the change in terms of the initial specific volume which can be used to calculate the change of volume once the initial specific volume is known.

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Most popular questions from this chapter

During some actual expansion and compression processes in piston-cylinder devices, the gases have been observed to satisfy the relationship \(P V^{n}=C\) where \(n\) and \(C\) are constants. Calculate the work done when a gas expands from \(350 \mathrm{kPa}\) and \(0.03 \mathrm{m}^{3}\) to a final volume of \(0.2 \mathrm{m}^{3}\) for the case of \(n=1.5\).

A \(0.5-m^{3}\) rigid tank contains nitrogen gas at \(600 \mathrm{kPa}\) and \(300 \mathrm{K}\). Now the gas is compressed isothermally to a volume of \(0.1 \mathrm{m}^{3} .\) The work done on the gas during this compression process is \((a) 720 \mathrm{kJ}\) (b) \(483 \mathrm{kJ}\) \((c) 240 \mathrm{kJ}\) \((d) 175 \mathrm{kJ}\) \((e) 143 \mathrm{kJ}\)

An insulated rigid tank is divided into two compartments of different volumes. Initially, each compartment contains the same ideal gas at identical pressure but at different temperatures and masses. The wall separating the two compartments is removed and the two gases are allowed to mix. Assuming constant specific heats, find the simplest expression for the mixture temperature written in the form $$T_{3}=f\left(\frac{m_{1}}{m_{3}}, \frac{m_{2}}{m_{3}}, T_{1}, T_{2}\right)$$ where \(m_{3}\) and \(T_{3}\) are the mass and temperature of the final mixture, respectively.

A spring-loaded piston-cylinder device contains \(1 \mathrm{kg}\) of carbon dioxide. This system is heated from 100 kPa and \(25^{\circ} \mathrm{C}\) to \(1000 \mathrm{kPa}\) and \(300^{\circ} \mathrm{C} .\) Determine the total heat transfer to and work produced by this system.

A well-sealed room contains \(60 \mathrm{kg}\) of air at \(200 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). Now solar energy enters the room at an average rate of \(0.8 \mathrm{kJ} / \mathrm{s}\) while a \(120-\mathrm{W}\) fan is turned on to circulate the air in the room. If heat transfer through the walls is negligible, the air temperature in the room in 30 min will be \((a) 25.6^{\circ} \mathrm{C}\) (b) \(49.8^{\circ} \mathrm{C}\) \((c) 53.4^{\circ} \mathrm{C}\) \((d) 52.5^{\circ} \mathrm{C}\) \((e) 63.4^{\circ} \mathrm{C}\)

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