Chemistry: Ten kilograms of R-134a fill a 1.595-m3 weighted piston-cylinder device at a temperature of −26.4°C. The container is now heated until the temperature is 100°C. Determine the final volume of the R-134a.

a || b4x + 72 + 2x - 12 = 1806x + 60 = 1806x = 180 -606x = 120x = 120/6x = 20...

Ten kilograms of R-134a fill a 1.595-m3 weighted piston-cylinder

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20.01.2022

Explanation:

The given data is as follows.

Mass of refrigerant, m = 10 kg

Volume of the refrigerant, V = 1.595 $m^{3}$

Formula for specific volume of the refrigerant is as follows.

v = $\frac{V}{m}$

= 0.1595 $m^{3}/kg$

So, at $-26.4^{o}C$ specific volume will be within $v_{f}$ and $v_{g}$ and pressure is constant.

The fluid will be in super-heated state at temperature $100^{o}C$ and at T = $-26.4^{o}C$ pressure 1 bar = 0.1 MPa.

According to super-heated tables, the specific volume is v = 0.30138 $m^{3}/kg$ .

Hence, the final volume will be calculated as follows.

$V_{f} = v \times m$

= $0.30138 m^{3}/kg \times 10 kg$

= 3.0138 $m^{3}$

Thus, we can conclude that final volume of the R-134a is 3.0138 $m^{3}$ .

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