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2 years ago

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Calculate the final temperature of 12.0 g of Argon (considered as an ideal gas) that is expanded reversibly and adiabatically fr

om V1 = 1.00 dm3 at T1 = 273.15 K to V2 = 3.00 dm3 . CV,m(Ar) = (3/2) * R.

1 answer:

Deffense [45]2 years ago

3 0

Answer:

Explanation:

For adiabatic change the expression is

= $PV^\gamma=constant$

$=(\frac{RT}{V})V^\gamma = constant$

$=TV^{\gamma-1} = constant$

$=T_1V_1^{\gamma-1} =$ $=T_2V_2^{\gamma-1}$

for Argon γ = 1.67

273.15 x 1 = T₂ x $3^{1.67-1}$

T₂ = 273.15 / $3^{0.67}$

= 273.15/ 2.0877

= 130.83 K.

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Mass of solution = Mass of ethylene glycol + Mass of water

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Now we have to calculate the volume of solution.

$\text{Volume of solution}=\frac{\text{Mass of solution}}{\text{Density of solution}}=\frac{2.114g}{1.070g/mL}=1.975mL$

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$\text{Volume percent}=\frac{\text{Volume of ethylene glycol}}{\text{Volume of solution}}\times 100=\frac{1mL}{1.975mL}\times 100=50.63\%$

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(d) Now we have to calculate the molality.

$\text{Molality}=\frac{\text{Mass of ethylene glycol}\times 1000}{\text{Molar mass of ethylene glycol}\times \text{Mass of water (in g)}}$

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