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malfutka [58]
3 years ago
8

Help me on a and b pls

Chemistry
1 answer:
marusya05 [52]3 years ago
5 0

Answer:

yes, it is increased by atomic mass

Explanation:

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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 intern
Arada [10]

<u>Answer:</u> The volume of the vessel is 0.1542m^3 and total internal energy is 162.0 kJ.

<u>Explanation:</u>

  • 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.

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