Expand your Mind, and Think of an Ideal Gas - Solutions

p.413, 12, 13, 34, 35, 38, 40 (What is V2/V1?)

12. Notice that you can do this problem without converting to meters, since you end up dividing delta_L by L0, which divides out the units anyway.

You should be able to find that delta_T = delta_L/(alpha*L0) = -89 degrees. (Remember change is final minus initial!) Adding this to the initial temperature of 20 degrees Celsius gives the final temperature of -69 degrees.

13. The key here is multiplying the final lengths of the plate after expansion.

(L + dL)(W + dW) = LW + W*dL + L*dW + dL*dW

As I mentioned in class, the last term is a higher order term, and can be neglected.

If we then substitute using the thermal expansion equation for dW = alpha*W*dT and dL = alpha*L*dT, we get that the final area is LW + 2*alpha*LW*dT.

The question asks for the change in area, so we just have to subtract the initial area LW to get what the text says should be the answer, 2*alpha*L*W*dT.

34. The first task is to determine the number of moles of the gas. Since there are 18.5 kg (18,500 g) of Nitrogen, and the molar weight of Nitrogen gas is 28 g/mol, we can divide to get that there are 661 moles of N2 gas in the tank.

We now have enough information to solve for V = nRT/P (remember that standard temperature and pressure means 273 K, 1.013 x 10⁵ Pa)

V = 14.8 m^3.

For part b, the tank has not changed, so it has the same volume as you found for the first part. We do know there is a new number of moles in the tank (661 + additional 536 moles).

Since we are looking for pressure, we again must solve for P in the ideal gas equation.

P = nRT/V = 1.84 x 10⁵ Pa

35. Remember we have to use absolute pressure, so the initial pressure in the container is (1 atm + .35 atm)*(1.013 x 10⁵ Pa/atm) = 1.37 x 10⁵ Pa

Solving, V = nRT/P = 0.439 m^3

b) Here we are changing states for the gas, so we need to use the division trick from class.

P1V1 = nRT1 and P2V2 = nRT2 , which can be combined to give
(P2/P1)(V2/V1) = T2/T1

Solving, T2 = 210 K = -63 degrees Celsius.

38. IN this case, since it is a change of state problem, we do NOT have to convert the pressure to Pascals before doing anything. We can use the same division method as before.

(P2/P1)(V2/V1) = T2/T1

P2 = (T2/T1)(V1/V2)*P1

P2 = (323.2 K/291.2 K)*(55.0 L/48.8 L)* 1 atm = 3.03 atm

40. Again, using the same method:
(P2/P1)(V2/V1) = T2/T1

(V2/V1) = (T2/T1)*(P1/P2) = 1.4