During which two phases would surfer’s most likely benefit
2 answers:
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Answer:
The total pressure of the solution is 598.66 torr
while the vapor pressure of the components is vapor pressure of benzene is =502 torr and that of toluene = 96.6 tor
Explanation:
To solve this question, we need to look at the known variables, the unknown variables, and then we select the appropriate relations between the known variables and the unknown
Here we have the temperature of the gases t = 80°C
the pressures of benzene = 753 torr
the pressures of toluene = 290 torr
mole fraction of benzene in solution = 2/3
mole fraction of toluene in solution = 1/3
the unknown variables = composition of the solution and the total pressure of the solution
From here given the known variables and the required variables, the appropriate relation bstween the known and unknown variables is Route's Law
Raoult's relates the vapor pressure of a mixture of gases to the mole fraction of solute gases introduced in the gas solution. Raoult's Law is expressed by the formula: Psolution = Χsolvent1×Psolvent1 + Χsolvent2×Psolvent2+ ... .
where. Psolution = vapor pressure of the solution and
Χsolvent1 = mole fraction of solvent 1
Psolvent1 = vapor pressure of solvent1
Thus we have by Raoult's Law
=
×
+
×
which is =2/3×753 + 1/3×290 = 502 + 96.6 = 598.66 torr
and composition vapor pressure of benzene is 502 torr
while the composition vapor pressure of toluene is 96.6 torr
Answer:
a) The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules, b) The area of the region enclosed by one loop of the curve is
.
Explanation:
a) The work, measured in joules, is a physical variable represented by the following integral:
Where
,
- Initial and final position, respectively, measured in meters.
- Force as a function of position, measured in newtons.
Given that and the fact that
when
, the spring constant (
), measured in newtons per meter, is:
Now, the work function is obtained:
The work required to stretch the spring from 20 centimeters to 25 centimeters is 0.313 joules.
b) Let be . The area of the region enclosed by one loop of the curve is given by the following integral:
By using trigonometrical identities, the integral is further simplified:
The area of the region enclosed by one loop of the curve is
.