A certain type of ideal gas at pressure p0 is contained in a plugged flask of volume V initially at temperature T1. The environmental pressure remains constant p0.

(A) As the flask is heated up to temperature T2, the plug pops out. What is the pressure p2 of the ideal gas inside of the flask right before the plug pops out?
(B)Following the preceding part, the plug is put back on once the pressure of the ideal gas becomes balanced with the environmental pressure p0, which is assumed to be constant. Remove the flask from heat and let it coll down to the initial temperature T1. What is the pressure p3 of the ideal gas in the flask now?
(C) Order p2,p3, and p0 in magnitude. Does it make sense? How to use this observation to tell whether there is loss of gas when the plug pops out and to decide how much gas is lost?

To answer the questions (A)-(C), we can make use of the ideal gas law, which relates the pressure, volume, and temperature of an ideal gas. The ideal gas law is given by the equation:

PV = nRT

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature in Kelvin.

(A) In this case, the volume V and the number of moles n of the gas remain constant. The initial pressure p0, initial temperature T1, and final temperature T2 are given. We need to find the final pressure p2 before the plug pops out.

To solve for p2, we can use the ideal gas law, rearranged to solve for pressure:

P = (nRT) / V

Since n, R, and V are constant, we can write:

P1 / T1 = P2 / T2

Substituting the given values, we have:

p0 / T1 = p2 / T2

Solving for p2, we get:

p2 = (p0 * T2) / T1

(B) In this case, the plug is put back on once the pressure of the gas becomes balanced with the environmental pressure p0. The flask is then removed from heat and allowed to cool down to the initial temperature T1. We need to find the pressure p3 of the gas in the flask at this point.

Since the gas is now in equilibrium with the environmental pressure p0, the pressure inside the flask is equal to p0. Therefore, p3 = p0.

(C) To order the magnitudes of p2, p3, and p0, we can analyze their values:

- For p2, we found that p2 = (p0 * T2) / T1, where T2 > T1. Therefore, p2 > p0.
- For p3, we found that p3 = p0, which is equal to p0 itself.
- Since p2 > p0 and p3 = p0, we have p2 > p3.

It makes sense that p2 > p3 because heating the gas caused an increase in pressure (p2), while cooling the gas back to the initial temperature caused the pressure to return to the environmental pressure (p3).

To determine whether there is any loss of gas when the plug pops out and to decide how much gas is lost, we need more information. The ideal gas law does not give any insights into gas loss. However, if the volume or the number of moles of gas changes after the plug pops out, it could indicate gas loss. For example, if the volume increases or if the number of moles decreases, it suggests gas loss.

(A) To find the pressure p2 of the ideal gas inside the flask right before the plug pops out, we can use the ideal gas law:

PV = nRT

Where P is the pressure, V is the volume, n is the moles of gas, R is the ideal gas constant, and T is the temperature.

Since the volume V and the environmental pressure p0 remain constant, we can write the equation as:

p2V = nRT2

Where p2 is the pressure right before the plug pops out, T2 is the final temperature, and n is the moles of gas.

(B) After the plug pops out, the pressure of the ideal gas becomes balanced with the environmental pressure p0. Assuming the temperature is still T2, we can write:

p0V = nRT2

Where p0 is the environmental pressure and T2 is the final temperature.

To find the pressure p3 after the flask cools down to the initial temperature T1, we can use the equation:

p0V = nRT1

Where T1 is the initial temperature.

(C) To order the pressures p2, p3, and p0 in magnitude, we need to compare them.

From part (A), we have p2V = nRT2, and from part (B), we have p0V = nRT1.

Comparing these two equations, we can see that p2 is greater than p0 if T2 is greater than T1. This is because the temperature T2 is higher, which leads to a higher pressure p2 right before the plug pops out.

From part (B), we also know that after the flask cools down to the initial temperature T1, the pressure p3 will be equal to p0, as p0V = nRT1.

Therefore, the order of the pressures magnitude is: p3 = p0 < p2.

This observation can be used to tell if there is a loss of gas when the plug pops out. If p2 is greater than p0, it indicates that there might be a loss of gas. The difference between p2 and p0 can be used to estimate how much gas is lost, as it relates to the change in pressure due to the change in temperature.