In thermodynamic process, pressure of a fixed mass of a gas is changes in such a manner that the gas molecules gives out 20 J of heat and 10 J of work is done in the gas. If the initial internal energy of the gas was 40 J, then the final internal energy will be

Options

A. 30 J
B. 20 J
C. 60 J
D. 40 J

Explanation

To find the final internal energy of the gas after the thermodynamic process, we can use the first law of thermodynamics, which states:

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

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

Given:
- Heat released by the gas, \(\Delta Q = -20 \, \text{J}\) (since it is releasing heat),
- Work done on the gas, \(W = -10 \, \text{J}\) (since work is done on the gas),
- Initial internal energy, \(U_i = 40 \, \text{J}\).

### Step 1: Calculate the change in internal energy (\(\Delta U\))

Using the first law of thermodynamics:

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

Substituting the values:

\[ \Delta U = (-20 \, \text{J}) - (-10 \, \text{J}) \]

This simplifies to:

\[ \Delta U = -20 \, \text{J} + 10 \, \text{J} = -10 \, \text{J} \]

### Step 2: Find the final internal energy (\(U_f\))

The change in internal energy is defined as:

\[ \Delta U = U_f - U_i \]

Rearranging this gives:

\[ U_f = U_i + \Delta U \]

Substituting the known values:

\[ U_f = 40 \, \text{J} + (-10 \, \text{J}) \]

This simplifies to:

\[ U_f = 40 \, \text{J} - 10 \, \text{J} = 30 \, \text{J} \]

### Final Answer:
The final internal energy of the gas is \(U_f = 30 \, \text{J}\).

---