A .02 kg of R-134a fills a 0.14-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 R-134a. Use data from the refrigerant tables.

The answer is 0.06027 m³

Explanation:

Solution

Given that:

The first step to take is to determine the initial state of the volume for R-134a refrigerant

Now,

v₁ =V/m

V = the volume of weighted piston cylinder device at the normal state

m = the mass of the R-134a refrigerant

Thus,

We substitute the values 0.14 for V and 0.2 kg for m

Which results in

v₁ = 0.14/0.2

v₁  = 0.7 m₃/kg

The next step is to find the saturated pressure of the R-134a refrigerant from the temperature table of saturated refrigerant R-134a which is equivalent to the normal temperature of 26.4°C.

Thus, by applying the method of interpolation we have the following

P₁ =101.73 - ((101.73-92.76) * (-26-(-26.4)/-26- (-28))

P₁ = 99.936 kPa

So,

The refrigerant in the weighted piston–cylinder device is then heated until the temperature gets to a 100°C

Hence, the temperature and pressure at a state of two becomes

P₂ = 99.936 kPa

T₂ = 100°C

The next step is to determine the specific volume of the refrigerant R-134a at a final state from the super heated refrigerant R-134a which is equivalent to the pressure of 99.936 kPa

v₂ =0.30138 m³ /kg

Now,

we now calculate the final state of  the weighted piston cylinder device

V₂ = mv₂

V₂ = 0.2 * 0.30138

V₂ = 0.06027 m³

Hence ,the final volume of the weighted cylinder piston device is 0.06027 m³

it was a fall down when it happens

explanation:

answer:

6.3 is the answer

explanation:

because it is the ex