In this example we see how thermal expansion can actually throw off the accuracy of a length-measuring device—namely, a tape mea
sure. A surveyor uses a steel measuring tape that is exactly 50.000 mm long at a temperature of 20 ∘C∘C. What is its length on a hot summer day when the temperature is 35 ∘C∘C?
2 answers:
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Answer: high temperature and low pressure
Explanation:
The Ideal Gas equation is:
Where:
is the pressure of the gas
is the volume of the gas
the number of moles of gas
is the gas constant
is the absolute temperature of the gas in Kelvin
According to this law, molecules in gaseous state do not exert any force among them (attraction or repulsion) and the volume of these molecules is small, therefore negligible in comparison with the volume of the container that contains them.
Now, real gases can behave approximately to an ideal gas, under the conditions described above and taking into account the following:
When <u>temperature is high</u> a real gas approximates to ideal gas, because the molecules move quickly, preventing the repulsion or attraction forces to take effect. In addition, at <u>low pressures</u>, the volume of molecules is negligible.
1)
The electrostatic force between two charges is given by
(1)
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the separation between the charges
When the two spheres are brought in contact with each other, the charge equally redistribute among the two spheres, such that each sphere will have a charge of
where Q is the total charge between the two spheres.
So we can actually rewrite the force as
And since we know that
r = 1.41 m (distance between the spheres)
(the sign is positive since the charges repel each other)
We can solve the equation for Q:
So, the final charge on the sphere on the right is
2)
Now we know the total charge initially on the two spheres. Moreover, at the beginning we know that
(we put a negative sign since the force is attractive, which means that the charges have opposite signs)
r = 1.41 m is the separation between the charges
And also,
So we can rewrite eq.(1) as
Solving for q1,
Since , we can substituting all numbers into the equation:
which gives two solutions:
Which correspond to the values of the two charges. Therefore, the initial charge q1 on the first sphere is