A gas receives an amount of heat equal to 110 J and performs 40 J of work. The change in the internal energy of the gas is

Options

A. 70 J
B. 150 J
C. 110 J
D. 40 J

Explanation

To solve the problem, we will use the first law of thermodynamics, which states:

\[ \Delta Q = \Delta U + \Delta W \]

Where:
- \(\Delta Q\) is the heat added to the system,
- \(\Delta U\) is the change in internal energy,
- \(\Delta W\) is the work done by the system.

Given:
- \(\Delta Q = 110 \, \text{J}\) (the amount of heat received by the gas),
- \(\Delta W = 40 \, \text{J}\) (the work done by the gas).

We need to find the change in internal energy \(\Delta U\).

### Step 1: Rearranging the equation
From the first law of thermodynamics, we can rearrange the equation to solve for \(\Delta U\):

\[ \Delta U = \Delta Q - \Delta W \]

### Step 2: Substituting the known values
Now, we can substitute the known values into the equation:

\[ \Delta U = 110 \, \text{J} - 40 \, \text{J} \]

### Step 3: Performing the calculation
Now, we perform the subtraction:

\[ \Delta U = 70 \, \text{J} \]

### Conclusion
The change in the internal energy of the gas is:

\[ \Delta U = 70 \, \text{J} \]