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vagabundo [1.1K]
2 years ago
12

When the piston in the Boyle’s Law apparatus is at rest, what is the relationship between the pressure of the trapped gas and th

e pressure on the outside of the piston?
Chemistry
2 answers:
Andru [333]2 years ago
7 0

Answer:

Pressure of the trapped gas and the pressure on the outside on the piston are the same.

Explanation:

When the piston is at rest, then it exist a mechanical equilibrium, that is to say, that pressure of the trapped gas is equal to the pressure on the outside of the piston.

salantis [7]2 years ago
5 0

Answer:

Check the last paragraph in the Explanation section.

Explanation:

In order to understand the Behavior of gases that is how ideal gases and real gases behave, scientists proposed several laws and one of them is the Boyle’s Law. The Boyle’s Law states that at a constant temperature the pressure of a given mass of an ideal gas is inversely proportional to its volume. The Boyle’s Law can be represented mathematically by using the equation below;

PV = k. Where P= pressure, V = volume and k = constant.

Hence, P1V1 = P2V2.

When the piston in the Boyle’s Law apparatus is at rest, the area of the Piston would be multiplied by the pressure of the trapped gas which will give the balanced weight. It will also balanced the additional weight.

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One mole of an ideal gas with a volume of 1.0 L and a pressure of 5.0 atm is allowed to expand isothermally into an evacuated bu
Deffense [45]

Answer:

w= - 1.7173 kJ, q= 1.7173 kJ, q(rev) = 1717.3 J = 1.7173 kJ.

Explanation:

Okay, from the question we are given the information below;

Number of moles, n= 1 mole; initial volume, v(1) = 1.0 litres (L); pressure (p) = 5atm, final volume(v2) = 2.0 Litres(L) ; the workdone, w= not given; the heat, q and q(rev)= not given and the gas was said to expand isothermally.

So, this question is a question from the part of chemistry known as thermodynamics. Therefore, grip yourself we are delving into thermodynamics 'waters' now.

For expansion isothermally; the workdone, w= -nRT ln v2/v1.

Where T= temperature= 25° C = 298 k and R= gas constant.

Therefore; workdone, w = - 1 × 8.314 × 298 × ln(2/1).

Workdone,w= - 1717.32204643. =

- 1717.3 Joules (J).

==> Workdone,w= - 1.7173 kJ.

Then, we are to find q. q can be solved by using the first law of thermodynamics, which by mathematical representation is:

∆U= q + w. Where ∆U= change in internal enegy. Since the question is dealing with isothermal expansion, there is this rule that says for an isothermal expansion ∆U = 0.

Hence, 0 =q + [- 1717.3 Joules (J)].

q=1717.3 J = 1.7173 kJ.

Finally, the q(rev) which is= nRT ln (v2/V1).

q(rev) = 1 × 8.314 × 298 ln (2/1).

q(rev) = 1717.3 J = 1.7173 kJ.

PS: please note the negative signs in the workdone and the positive sign in the q(rev).

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