A mixture of gas expands from 0.01 m³ to 0.06 m³ at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. What is the change in internal energy of the mixture? A mixture of gas expands from 0.01 m³ to 0.06 m³ at a constant pressure of 1 MPa and absorbs 84 kJ of heat during the process. What is the change in internal energy of the mixture? Read more

Steps to Solve

1. State the First Law of Thermodynamics

The first law of thermodynamics states that the change in internal energy ($\Delta U$) of a system is equal to the heat added to the system ($Q$) minus the work done by the system ($W$).

$$ \Delta U = Q - W $$

2. Calculate the work done during the expansion

Since the expansion occurs at constant pressure ($P$), the work done by the gas is given by:

$$ W = P \Delta V = P (V_2 - V_1) $$

where $V_1$ is the initial volume and $V_2$ is the final volume. Given $V_1 = 2 \text{ m}^3$, $V_2 = 6 \text{ m}^3$, and $P = 200 \text{ kPa} = 200 \times 10^3 \text{ Pa}$, we have:

$$ W = (200 \times 10^3 \text{ Pa})(6 \text{ m}^3 - 2 \text{ m}^3) = (200 \times 10^3 \text{ Pa})(4 \text{ m}^3) = 800 \times 10^3 \text{ J} = 800 \text{ kJ} $$

3. Calculate the change in internal energy

The heat absorbed by the gas is $Q = 900 \text{ kJ}$. Using the first law of thermodynamics:

$$ \Delta U = Q - W = 900 \text{ kJ} - 800 \text{ kJ} = 100 \text{ kJ} $$