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Doss [256]
3 years ago
6

one kilogram of water (V1 = 1003 cm^3*kg-1) in a piston cylinder device at (25°C) and 1 bar is compressed in a mechanically reve

rsible, isothermal process to 1500 bar. Determine Q , W, ΔU , ΔH , and ΔS given that β = 250 x 10-6K-1 and K = 45 x 10^-6 bar-1.
Chemistry
1 answer:
vova2212 [387]3 years ago
4 0

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

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Answer:

12.44 g

Explanation:

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n(C4H10) = m(C4H10)/M(C4H10) = 4.1 / 58g/mol = 0.0707 mol (excess).

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Since the ratio of O2 to octane is 13 : 2 we can divide 0.0707 by 2 to get 0.03535 and divide 0.809 by 13 to get 0.062.

mass of CO2 produced =

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M = 0.5656/2 * 44

M = 0.2828 * 44

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3 years ago
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Answer:

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2 years ago
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A 25.0 g bold made of an alloy absorbed 250 J of heat as its temperature changed from 25.0 °C to 78.0 °C. What is the specific h
nata0808 [166]

Answer:

Specific heat of alloy = 0.2 j/ g.°C

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Given data:

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Heat absorbed = 250 J

Initial Temperature = 25°C

Final temperature = 78°C

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Change in temperature:

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ΔT = 53°C

Now we will put the values in formula.

Q = m.c. ΔT

250 j = 25 g × c ×53°C

250 j = 1325 g.°C × c

250 j / 1325 g.°C = c

c = 0.2 j/ g.°C

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