Question

A rigid vessel contains 3.98 kg of refrigerant-134a at 700 kPa and 60°C. Determine the volume of the vessel and the total internal energy. m3 (Round to four decimal places) kJ (Round to one decimal place)

Answer 1

Answer: The volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.

Explanation:

• To calculate the volume of water, we use the equation given by ideal gas, which is:

$PV=nRT$

or,

$PV=\frac{m}{M}RT$

where,

P = pressure of container = 700 kPa

V = volume of container = ? L

m = Given mass of R-134a = 3.98 kg = 3980 g    (Conversion factor: 1kg = 1000 g)

M = Molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.31\text{L kPa }mol^{-1}K^{-1}$

T = temperature of container = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$700kPa\times V=\frac{3980g}{102.03g/mol}\times 8.31\text{L kPa }\times 333K\\\\V=154.21L$

Converting this value into $m^3$, we use the conversion factor:

$1m^3=1000L$

So, $\Rightarrow (\frac{1m^3}{1000L})\times 154.21L$

$\Rightarrow 0.1542m^3$

• To calculate the internal energy, we use the equation:

$U=\frac{3}{2}nRT$

or,

$U=\frac{3}{2}\frac{m}{M}RT$

where,

U = total internal energy

m = given mass of R-134a = 3.98 kg = 3980 g  (Conversion factor: 1kg = 1000g)

M = molar mass of R-134a = 102.03 g/mol

R = Gas constant = $8.314J/K.mol$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$U=\frac{3}{2}\times \frac{3980g}{102.03g/mol}\times 8.314J/K.mol\times 333K\\\\U=161994.6J$

Converting this into kilo joules, we use the conversion factor:

1 kJ = 1000 J

So, 161994.6 J = 162.0 kJ

Hence, the volume of the vessel is $0.1542m^3$ and total internal energy is 162.0 kJ.