Answer:
Q = -18118.5KJ
W = -18118.5KJ
∆U = 0
∆H = 0
∆S = -60.80KJ/KgK
Explanation:
W = RTln(P1/P2)
P1 = 1bar = 100KN/m^2, P2 = 1500bar = 1500×100 = 150000KN/m^2, T = 23°C = 23 + 273K = 298K
W = 8.314×298ln(100/150000) = 8.314×298×-7.313 = -18118.5KJ ( work is negative because the isothermal process involves compression)
∆U = Cv(T2 - T1)
For an isothermal process, temperature is constant, so T2 = T1
∆U = Cv(T1 - T1) = Cv × 0 = 0
Q = ∆U + W = 0 + (-18118.5) = 0 - 18118.5 = -18118.5KJ
∆H = Cp(T2 - T1)
T2 = T1
∆H = Cp(T1 - T1) = Cp × 0 = 0
∆S = Q/T
Mass of water = 1kg
Heat transferred (Q) per kilogram of water = -18118.5KJ/Kg
∆S = (-18118.5KJ/Kg)/298K = -60.80KJ/KgK
Answer:
The correct answer is option 2.
Explanation:
Colligative property is defined as property which depends upon only on the numbers of particles of solute dissolved in definite amount of solvent, It do not depend on the nature of the solute.
For example : NaCl solution with 0.4 molal will show same colligative properties as a that of the glucose solution with 0.04 molal concentration.
The following are the examples of colligative property:
1. Relative lowering of vapor pressure.
2. Osmotic pressure
3. Elevation in boiling points
4. Depression in freezing point
Explanation:
if we fix the temperature, we are just left with PV = constant for the gas law. So, in this situation, if the volume is doubled, the pressure must go down by one-half. And vice-versa. The simplest illustration of this would be a cylinder with a plunger on one end: if you push the plunger in so that the volume of the cylinder is halved and the temperature remains constant, then the pressure will double.
Answer:
This is a precipitation reaction: AgCl is the formed precipitate.
Let A be the 80% solution and B be the 20% solution and P be the produce solution of 70%. Va and Vb and Vp are the volumes of A and B and P respectively.
Va + 60 = Vp
0.7Vp = 0.8Va + 0.2(60)
Substituting the value of Vp from the first equation:
0.7(Va + 60) = 0.8Va + 12
30 = 0.1Va
Va = 300 gallons
