Question

55 mol of an ideal gas with CV,m=3R/2 undergoes the transformations described in the following list from an initial state described by T=365K and P=7.00bar.

1. The gas undergoes a reversible adiabatic expansion until the final pressure is one-fourth its initial value.

2. The gas undergoes an adiabatic expansion against a constant external pressure of 1.75 bar until the final pressure is one-fourth its initial value.

What is the net heat exchanged by the system after step 1 and 2?

Answer 1

The net heat exchanged by the system after step 1 and step 2 is zero.

Step-by-step explanation:

First, let's calculate the final pressure after step 1. In an adiabatic expansion, the relationship between pressure and volume is given by P₁V₁^y = P₂V₂^y, where P₁ and V₁ are the initial pressure and volume, P₂ and V₂ are the final pressure and volume, and y is the heat capacity ratio (CV,m/Cp,m) of the gas.

Plugging in the given values, we have (7.00 bar)(2.5)^y = (1/4)(7.00 bar)(2.0)^y. Since CV,m = 3R/2, which is the heat capacity ratio for a monatomic ideal gas, we can substitute y = 5/3.

Simplifying the equation, we get (2.5)^y = (1/4)(2.0)^y. Taking the natural logarithm of both sides and solving for y, we find y ≈ 0.693. Substituting this value for y into the equation, we can solve for P₂, which gives us P₂ ≈ 1.70 bar.

Now, let's move on to step 2. In an adiabatic process against constant external pressure, the relationship between pressure and volume is given by P₁V₁^y = P₂V₂^y. Plugging in the values, we have (1.70 bar)(2.0)^y = (1/4)(1.75 bar)(2.0)^y. Solving for y, we find y = 0.

Since y = 0, the gas is undergoing an isothermal process. In an isothermal process, the heat exchanged by the system is zero.

Therefore, the net heat exchanged by the system after step 1 and step 2 is zero.