Question

A balloon filled with 46.6 mol helium has a volume of 1044. l at 0.0°C and 1.00 atm pressure. The temperature of the balloon is increased to 48.0°C as it expands to a volume of 1228 l, the pressure remaining constant. Calculate q, w, and Δe for the helium in the balloon. (The molar heat capacity for helium gas is 20.8 J/°C·mol.)

Answer 1

To calculate q, w, and Δe for the helium in the balloon, we use the first law of thermodynamics. q can be calculated as nCΔT, where n is the number of moles, C is the molar heat capacity, and ΔT is the change in temperature. The change in internal energy (Δe) is equal to q, as the work done on or by the system is zero (w = 0) due to constant pressure.

Step-by-step explanation:

To calculate q, w, and Δe for the helium in the balloon, we can use the first law of thermodynamics, which states that Δe = q + w. Since the pressure remains constant, the work done on or by the system is zero (w = 0). Therefore, we only need to calculate the heat transferred (q) and the change in internal energy (Δe).

To calculate q, we can use the equation q = nCΔT, where n is the number of moles, C is the molar heat capacity, and ΔT is the change in temperature. In this case, n = 46.6 mol, C = 20.8 J/°C·mol, and ΔT = (48.0°C - 0.0°C) = 48.0°C.

Substituting these values into the equation, we get q = (46.6 mol)(20.8 J/°C·mol)(48.0°C) = 46,049.28 J.

Since the pressure remains constant, the change in internal energy can be calculated as Δe = q = 46,049.28 J.