Complete question:
ΔU for a van der Waals gas increases by 475 J in an expansion process, and the magnitude of w is 93.0 J. calculate the magnitude of q for the process.
Answer:
The magnitude of q for the process 568 J.
Explanation:
Given;
change in internal energy of the gas, ΔU = 475 J
work done by the gas, w = 93 J
heat added to the system, = q
During gas expansion process, heat is added to the gas.
Apply the first law of thermodynamic to determine the magnitude of heat added to the gas.
ΔU = q - w
q = ΔU + w
q = 475 J + 93 J
q = 568 J
Therefore, the magnitude of q for the process 568 J.
The question is incomplete. The complete question is:
At 25◦C and atmospheric pressure the volume change of mixing of binary liquid mixtures of species 1 and 2 is given by the equation:
ΔV = x1x2(45x1 + 25x2)
Where ΔV is in cm3-mol-1. At these conditions, the molar volumes of pure liquid 1 and 2 are V1= 110 and V2= 90 cm3-mol-1. Determine the partial molar volumes 1VE and 2VE in a mixture containing 40 mole percent of specie 1.
Answer:
1VE = 117.92 cm³.mol⁻¹, 2VE = 97.92 cm³.mol⁻¹
Explanation:
In the equation given, x represents the molar fraction of each substance, thus x1 = 0.4 and x2 = 0.6. Because of the mixture, the molar partial volume of each substance will change by a same amount, which will be:
ΔV = 0.4*0.6(45*0.4+ 25*0.6)
ΔV = 7.92 cm³.mol⁻¹
1VE - V1 = 7.92
1VE = 7.92 + 110
1VE = 117.92 cm³.mol⁻¹
2VE - V2 = 7.92
2VE = 7.92 + 90
2VE = 97.92 cm³.mol⁻¹
14kg=Number of Apples*.2kg
70 Apples *8 Apples each=560 seeds
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