Steam is contained in a closed rigid container with a volume of 1 m3. initially, the pressure and temperature of the steam are 7 bar and 500°c, respectively. the temperature drops as a result of heat transfer to the surroundings. determine

Question:

1. Steam is contained in a closed rigid container with a volume of 1 m³. Initially, the pressure and temperature of the steam are 7 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine

a) the temperature at which condensation first occurs, in °C,

b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.

c) What is the volume, in m³, occupied by saturated liquid at the final state?

The answers to the question are;

a) The temperature at which condensation first occurs is 163.056 °C.

b) The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.844.

c) The volume, occupied by saturated liquid at the final state is 1.66×10⁻³ m³.

Explanation:

To solve the question, we note that we are required check the steam tables for the properties of steam.

(a) From P×V = n×R×T

we find the number of moles as

n = P₁V₁/(RT₁)

Where

n = Number of moles

P₁ = Initial pressure = 7 bar = 700000 Pa

V₁ = Volume  = 1 m³

T₁ = Temperature = 500 C = 773.15 K

R = Universal gas constant = 8.31451  J/(gmol·K)

Therefore n  = (700000 Pa×1 m³)/(8.31451  J/(gmol·K)×773.15 K)

n = 108.89 moles

Molar mass of H₂O = 18.01528 g/mol

Therefore initial mass of steam = Number of moles  × Molar mass

= 1961.73 g

From steam tables, condensation first start to occur at

151.836 °C + ((179.886 °C - 151.836 °C)/(10 bar - 5 bar))×(2 bar) = 163.056 °C

Therefore condensation first start to occur at 163.056 °C.

b) At 0.5 bar we have

n = P₂V₂/(RT₂)

Where

V₂ = V₁  = Volume = 1 m³

P₂ = Pressure = 0.5 bar = 50000 Pa

T₂ = Saturation temperature at 0.5 Pa = 81.3167 °C  = 354.4667 K

R = Universal gas constant = 8.31451  J/(gmol·K)

n = (50000 Pa × 1 m³)/(354.4667 K×8.31451  J/(gmol·K)) = 16.965 moles

Mass of steam left at 0.5 bar = 16.965 moles × 18.01528 g/mol = 305.632 g.

Mass of condensed steam

= Initial mass of steam at 7 bar - Final mass of steam at 0.5 bar

= 1961.73 g - 305.632 g =  1656.095 g

The fraction of the total mass that has condensed at 0.5 bar is given by

Mass fraction of condensed steam

= (Mass of condensed steam)/(Initial mass of steam)

1656.095 g/1961.73 g = 0.844.

c) The volume occupied by saturated liquid at the final state is given by

Volume = Mass/Density = (1656.095 g)/(997 kg/m³)

= (1.656095 kg)/(997 kg/m³) = 1.66×10⁻³ m³.

answer is: c. the experiment shows that the red substance experienced a chemical change.

this is chemical change (chemical reaction), because two new substances are formed (metalig liquid and gas), the atoms are rearranged and the reaction is followed by an energy change.

this is example of chemical decomposition.

chemical decomposition (analysis) is the separation of a single chemical compound into its two or more elemental parts or to simpler compounds.  

chemical decomposition is opposite of chemical synthesis.

the answer is   c.) 1575 j.

explanation: