The answer is
d. To what temperature can the steam at 80 bar be cooled before a liquid will appear?
Answer:
10.5L
Explanation:
The volume in this question can be calculated by using the formula for gas law equation as follows:
PV = nRT
Where;
P = pressure (atm)
V = volume (litres)
n = number of moles (mol)
R = gas law constant (0.0821 Latm/molK)
T = temperature (K)
According to the provided information in this question:
P = 6.18 atm
V = ?
n = 2.35 moles
R = 0.0821 Latm/molK
T = 63°C = 63 + 273 = 336K
Using PV = nRT
V = nRT/P
V = 2.35 × 0.0821 × 336/ 6.18
V = 64.83/6.18
V = 10.49
V = 10.5L
Answer:
It heats up.
Explanation:
Because thermal energy is based off the temperature of an object and the average kinetic energy of its particles increases. When the average kinetic energy of its particles increases, the object's thermal energy increases. Therefore, the thermal energy of an object increases as its temperature increase.
Answer:
There will many opportunities in the field of renewable energy science and engineering. It is an exciting field, with a lot of variety and room for new ideas.
Explanation:
This question is incomplete, the complete question is;
Tonksite is a solid at 300.00K. At 300.00 K its enthalpy of sublimation is 66.00 kJ/mol. The sublimation pressure at 300.00 K is 5.00 × 10⁻⁴ atm
Calculate the sublimation pressure of the solid at the melting point of 400.00 K assuming that the enthalpy of sublimation is not a function of temperature.
Answer: the sublimation pressure of the solid at the melting point is 0.3727 atm
Explanation:
Given that;
T1 = 300 K
T2 = 400 K
H_sub = 66 kJ/mol = 66000 J/mol
P1 = 5.00 × 10⁻⁴ atm
p2 = ?
now using the expression
log( p2 / 5.00 × 10⁻⁴ ) = (H_sub / R × 2.303 ) (( T2 - T1) / T1T2)
now we substitute of given values into the expression
log(p2/p1) = (66000 / 8.314 × 2.303 ) (( 400 - 300) / 300 × 400 )
p2 = 0.3727 atm
therefore the sublimation pressure of the solid at the melting point is 0.3727 atm
