Question

A monatomic ideal gas undergoes a quasi-static process that is described by the function p(V)=p₁+3(V−V₁), where the starting state is (p₁,V₁) and the final state (p₂,V₂). Assume the system consists of n moles of the gas in a container that can exchange heat with the environment and whose volume can change freely. (a) Evaluate the work done by the gas during the change in the state. (b) Find the change in internal energy of the gas. (c) Find the heat input to the gas during the change. (d) What are initial and final temperatures?

a) Work = 3/2 (p₂V₂ - p₁V₁), ΔU = 3/2 (p₂V₂ - p₁V₁), Q = 0, Tᵢⁿ^iᵗ^iᵃˡ = ? T_f^iⁿᵃˡ = ?
b) Work = 0, ΔU = 3/2 (p₂V₂ - p₁V₁), Q = 3/2 (p₂V₂ - p₁V₁), Tᵢⁿ^iᵗ^iᵃˡ = ? T_f^iⁿᵃˡ = ?
c) Work = 3/2 (p₂V₂ - p₁V₁), ΔU = 0, Q = 3/2 (p₂V₂ - p₁V₁), Tᵢⁿ^iᵗ^iᵃˡ = ? T_f^iⁿᵃˡ = ?
d) Work = 0, ΔU = 0, Q = 0, Tᵢⁿ^iᵗ^iᵃˡ = ? T_f^iⁿᵃˡ = ?

Answer 1

The work done by the gas is 3/2 (p₂V₂ - p₁V₁), the change in internal energy is also 3/2 (p₂V₂ - p₁V₁), there is no heat input, and the temperatures are unknown.

Step-by-step explanation:

In this problem, a monatomic ideal gas undergoes a quasi-static process described by the function p(V)=p₁+3(V−V₁). To evaluate the work done by the gas, we use the formula W = ∫p dV. Since the process is quasi-static, we can use the given function to find the limits of integration for the integral. The work done by the gas during the change in state is 3/2 (p₂V₂ - p₁V₁).

The change in internal energy of the gas can be found using the first law of thermodynamics, ΔU = Q - W. Since Q = 0 for this process, the change in internal energy is also 3/2 (p₂V₂ - p₁V₁).

Since Q = 0, there is no heat input to the gas during the change in state.

The initial and final temperatures cannot be determined from the given information.