A mixture of gas expands at a constant pressure, absorbing heat and increasing in volume. If the pressure is 1 MPa, the initial volume is 0.01 m, the final volume is 0.06 m, and th... A mixture of gas expands at a constant pressure, absorbing heat and increasing in volume. If the pressure is 1 MPa, the initial volume is 0.01 m, the final volume is 0.06 m, and the absorbed heat is 84 kJ, what is the change in internal energy of the mixture?
Understand the Problem
The question is asking us to calculate the change in internal energy of a gas mixture undergoing an expansion at constant pressure. We are given the initial and final volumes, the constant pressure, and the heat absorbed by the mixture. We can use the first law of thermodynamics to relate these quantities and solve for the change in internal energy.
Answer
$ \Delta U = -60 \, \text{kJ} $
Answer for screen readers
$$ \Delta U = -60 , \text{kJ} $$
Steps to Solve
- State the first law of thermodynamics
The first law of thermodynamics relates the change in internal energy ($\Delta U$) to the heat added to the system ($Q$) and the work done by the system ($W$):
$$ \Delta U = Q - W $$
- Calculate the work done during the expansion
Since the expansion occurs at constant pressure ($P$), the work done by the gas is given by:
$$ W = P \Delta V = P (V_f - V_i) $$
where $V_f$ is the final volume and $V_i$ is the initial volume. We are given that $P = 100 , \text{kPa} = 100 \times 10^3 , \text{Pa}$, $V_i = 1.0 , \text{m}^3$, and $V_f = 2.5 , \text{m}^3$. Therefore:
$$ W = (100 \times 10^3 , \text{Pa}) (2.5 , \text{m}^3 - 1.0 , \text{m}^3) = (100 \times 10^3 , \text{Pa}) (1.5 , \text{m}^3) = 150 \times 10^3 , \text{J} = 150 , \text{kJ} $$
- Calculate the change in internal energy
We are given that the heat absorbed by the mixture is $Q = 90 , \text{kJ}$. Using the first law of thermodynamics:
$$ \Delta U = Q - W = 90 , \text{kJ} - 150 , \text{kJ} = -60 , \text{kJ} $$
$$ \Delta U = -60 , \text{kJ} $$
More Information
The negative sign indicates that the internal energy of the gas mixture decreased during the expansion. This makes sense because the gas did work on the surroundings, and the heat absorbed was not enough to compensate for the energy lost as work.
Tips
A common mistake is to confuse the sign convention for work. In this case, since the gas is expanding, it is doing work on the surroundings, so the work done by the system is positive. If the problem stated the work done on the system, then the sign would be reversed. Another common mistake is failing to convert all values to SI units before plugging them into the equations.
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