Suppose you are asked to find the area of a rectangle that is 2.1-cm wide by 5.6-cm long. Your calculator answer would be 11.76
1 answer:
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Answer : The work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Explanation :
(a) At constant volume condition the entropy change of the gas is:
We know that,
The relation between the for an ideal gas are :
As we are given :
Now we have to calculate the entropy change of the gas.
(b) As we know that, the work done for isochoric (constant volume) is equal to zero.
(C) Heat during the process will be,
Therefore, the work, heat during the process and the change of entropy of the gas are, 0 J, 3333.003 J and -10 J respectively.
Step 1
The osmotic pressure is calculated as follows:
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Step 2
<em>Information provided:</em>
The mass of solute = 13.6 g
Volume of solution = 251 mL
Absolute temperature = T = 298 K
The molar mass of solute = M = 354.5 g/mol
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Step 3
Procedure:
1 L = 1000 mL => Volume = 251 mL x (1 L/1000 mL) = 0.251 L
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C = moles of solute/volume of solution (L)
C = mass of solute/(molar mass x Volume (L))
C = 13.6 g/(354.5 g/mol x 0.251 L)
C = 0.153 mol/L
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π = C x R x T
π = 0.153 mol/L x 0.082 atm L/mol K x 298 K
π = 3.74 atm
Answer: π = 3.74 atm