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DanielleElmas [232]
3 years ago
10

The volume of 160. g of CO initially at 273 K and 1.00 bar increases by a factor of two in different processes. Take CP,m to be

constant at the value 29.14 Jmol−1K−1 and assume ideal gas behavior. The temperature of the surroundings is 273 K.
A) Calculate ΔSsurroundings in an adiabatic reversible expansion.

B) Calculate ΔS in an adiabatic reversible expansion.

C) Calculate ΔStotal in an adiabatic reversible expansion.

D) Calculate ΔSsurroundings in an expansion against Pexternal = 0.

E) Calculate ΔS in an expansion against Pexternal = 0.

F) Calculate ΔStotal in an expansion against Pexternal = 0.

G) Calculate ΔSsurroundings in an isothermal reversible expansion.

H) Calculate ΔS in an isothermal reversible expansion.

I) Calculate ΔStotal in an isothermal reversible expansion.

Determine what processes are spontaneous.

1) adiabatic reversible expansion 2) expansion against Pexternal =0 3)isothermal reversible expansion
Chemistry
1 answer:
mel-nik [20]3 years ago
5 0

Answer:

Explanation:

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A D F

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When 0.49 g of a molecular compound was dissolved in 20.00 g of cyclohexane, the freezing point of the solution was lowered by 3
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Explanation:

Depression in freezing point:

\Delta T_f=i\times k_f\times \frac{w_2\times 1000}{M_2\times w_1}

where,

\Delta T_f = depression in freezing point  = 3.9^oC

k_f = freezing point constant  = 20.8^0C/m

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w_1 = mass of solvent (cyclohexane) = 20.00 g

M_2 = molar mass of solute = ?

Now put all the given values in the above formula, we get:

(3.9)^oC=1\times (20.8^oC/m)\times \frac{(0.49g)\times 1000}{M_2\times (20.00g)}

M_2=131g/mol

Therefore, the molar mass of solute is 131 g/mol

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