A frictionless piston-cylinder device contains air at 300 K and 1 bar and is heated until its volume doubles and the temperature reaches 600 K. Answer the following: A. You are interested in studying the air in the piston-cylinder device as a closed system. Draw a schematic of your device and the boundary that defines your system. Assume the cylinder is in horizontal position. B. Determine the final pressure of the air at the end of the process, in bar. Hint: use the ideal gas law equation. If you need the value for the universal gas constant ???????? ????in your textbook or in a chemistry book (or on-line). Just make sure your units are dimensionally correct. C. On a different occasion (different temperature and pressure), you find the piston-cylinder device contains 0.5 kmol of H2O occupying a volume of 0.009 m3. Determine the weight of the H2O in N. Hint: Start with the relationship between number of moles, molecular mass and mass. D. Determine the specific volume of the H2O (from Part C) in m3/kg.

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danialamin danialamin · Answer 1 · 2020-02-05T14:10:54+01:00

Answer:

Part a: The schematic diagram is attached.

Part b: The pressure at the end is 1 bar.

Part c: Weight of 0.5kmol of water is 88.2 N.

Part d: The specific volume is 0.001 m^3/kg

Explanation:

The schematic is given in the diagram attached.

Pressure is given using the ideal gas equation as

Here

[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\frac{1\times V_1}{300}=\frac{P_2\times 2V_1}{600}\\P_2=\frac{600}{600}\\P_2=1 bar[/tex]

So the pressure at the end is 1 bar.

Mass of 0.5kmol is given as follows

[tex]Mass=n_{moles} \times Molar \, Mass\\Mass=0.5 \times 10^3 \times 18 \times 10^{-3}\\Mass=9.0 kg[/tex]

Weight is given as

[tex]W=mxg\\W=9 \times 9.8\\W=88.2 \, N[/tex]

So weight of 0.5kmol of water is 88.2 N.

Specific volume is given as

[tex]v=\frac{Volume}{Mass}\\v=\frac{0.009}{9}\\v=0.001 m^3/kg[/tex]

So the specific volume is 0.001 m^3/kg