Question

A stoppered test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18OC. The cylindrically shaped stopper at the test tube's mouth has a diameter of 1.20 cm and will "pop off" the test tube if a net upward force of 11.5 N is applied to it.

To what temperature would one have to heat the trapped air in order to "pop off" the stopper? Assume the air surrounding the test tube is always at a pressure of 1.00 atm.

Answer 1

To determine the temperature at which the stopper will pop off, we can use the ideal gas law equation with the given information. By calculating the number of moles of gas and rearranging the equation, we find that the temperature needs to be heated to 1422 K in order to pop off the stopper.

Step-by-step explanation:

To determine the temperature at which the stopper will pop off, we need to use the ideal gas law equation: PV = nRT. In this case, the pressure (P) and volume (V) are constant, so the equation simplifies to nR(T2 - T1) = F, where n is the number of moles of gas, R is the ideal gas constant, T1 is the initial temperature, T2 is the final temperature, and F is the force applied to the stopper.

First, we need to find the number of moles of gas. We can use the equation n = PV / RT, where P is the pressure, V is the volume, and R is the ideal gas constant. Plugging in the given values: n = (1.00 atm)(0.0250 L) / (0.0821 L.atm/(mol.K))(273 + 18) = 0.00101 mol.

Next, we can rearrange the first equation to solve for T2: T2 = (F / (nR)) + T1. Plugging in the given values: T2 = (11.5 N / (0.00101 mol)(0.0821 L.atm/(mol.K))) + 18 = 1422 K.

Therefore, the temperature at which the trapped air needs to be heated in order to pop off the stopper is 1422 K.