Question

1/ A gas is compressed from 400 cm3 to 200 cm3 at a constant pressure of
100 kPa. At the same time, 100 J of heat energy is transferred out of the
gas. What is the change in internal energy of the gas during this process?

Answer 1

Answer on Question 67516, Physics, Molecular Physics, Thermodynamics

Question:

A gas is compressed from $400\, \text{cm}^3$ to $200\, \text{cm}^3$ at a constant pressure of $100\, \text{kPa}$. At the same time, $100\, \text{J}$ of heat energy is transferred out of the gas. What is the change in internal energy of the gas during this process?

Solution:

We can find the change in internal energy of the gas during this process from the first law of thermodynamics:

$$\Delta U = Q + W,$$

here, $\Delta U$ is the change in internal energy of the gas during this process, $Q = -100\, \text{J}$ is the heat energy that transferred out of the gas, $W$ is the work done by the gas.

By the definition of the work done by the gas, we get:

$$\begin{array}{l} W = -p \Delta V = -p \left(V_{\text{final}} - V_{\text{initial}}\right) = \\ = -100 \cdot 10^3\, \text{Pa} \cdot (200 \cdot 10^{-6}\, \text{m}^3 - 400 \cdot 10^{-6}\, \text{m}^3) = 20\, \text{J}. \end{array}$$

Finally, we can find the change in internal energy of the gas during this process:

$$\Delta U = Q + W = -100\, \text{J} + 20\, \text{J} = -80\, \text{J}.$$

Answer:

$$\Delta U = -80\, \text{J}.$$

Answer provided by https://www.AssignmentExpert.com