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Artist 52 [7]
3 years ago
5

What is the entropy change of the surroundings

Chemistry
1 answer:
KiRa [710]3 years ago
5 0

Answer: The entropy change of the surroundings will be -17.7 J/K mol.

Explanation: The enthalpy of vapourization for 1 mole of acetone is 31.3 kJ/mol

Amount of Acetone given = 10.8 g

Number of moles is calculated by using the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Molar mass of acetone = 58 g/mol

Number of moles = \frac{10.8}{58}=0.1862moles

If 1 mole of acetone has 32.3 kJ/mol of enthalpy, then

0.1862 moles will have = \frac{32.3}{1}\times 0.1862=5.828kJ/mol

To calculate the entropy change for the system, we use the formula:

\Delta S_{sys}=\frac{\Delta H_{vap}}{T(\text{ in K)}}

Temperature = 56.2°C = (273 + 56.2)K = 329.2K

Putting values in above equation, we get

\Delta S_{sys}=\frac{5.828}{329.2}=0.0177kJ/Kmol=17.7J/Kmol   (Conversion Factor: 1 kJ = 1000J)

At Boiling point, the liquid phase and gaseous phase of acetone are in equilibrium. Hence,

\Delta S_{system}+\Delta S_{surrounding}=0

\Delta S_{surrouding}=-\Delta S_{system}=-17.7J/Kmol

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An "empty" container is not really empty if it contains air. How may moles of nitrogen are in an "empty" two-liter cola bottle a
Lisa [10]

Answer:

1. 0.0637 moles of nitrogen.

2. The partial pressure of oxygen is 0.21 atm.  

Explanation:

1. If we assume ideal behaviour, we can use the Law of ideal gases to find the moles of nitrogen, considering that air composition is mainly nitrogen (78%), oxygen (21%) and argon (1%):  

V_{N_2}=V_{T}\times 0.78=2L \times 0.78 =1.56 L\\PV=nRT\\n_{N_2}=\frac{PV}{RT}=\frac{1 atm\times 1.56 L}{0.0821\frac{atmL}{molK}\times 298 K}\\n_{N_2}= 0.0637 mol

2. Now, in order to find he partial pressure of oxygen we need to find the total moles of air, and then the moles of oxygen. Then, we use these results to determine the molar fraction of oxygen, to multiply it with total pressure and get the partial pressure of oxygen as follows:

n_{total}=\frac{1 atm \times 2L}{0.0821 \frac{atmL}{molK}298K}=0.0817 mol

V_{O_2}=2L \times 0.21 = 0.42 L\\n_{O_2}=\frac {1atm \times 0.42 L}{0.0821 \frac{atm L}{mol K}298 K}=0.0172 mol\\X_{O_2}=\frac{n_{O_2}}{n_{total}}=\frac{0.0172 mol}{0.0817 mol}= 0.21

P_{O_2}=X_{O_2} \times P = 1 atm \times 0.21 = 0.21 atm

As you see, the molar fraction and volume fraction are the same because of the assumption of ideal behaviour.  

3 0
3 years ago
PLZZ HELP!!!
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How many moles of Ar gas are<br> present in a container with a<br> volume of 78.4 L at STP?
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<h3><u>Answer:</u></h3>

<u>1 mole of a gas at STP occupies 22.4 L volume </u>

<u>Now the volume is given =78.4 therefore,</u>

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