In a thermodynamic process pressure of a fixed mass of a gas is changed in such a manner that the gas releases 30 joules of heat and 10 joules of work was done on the gas. If the initial internal energy of the gas was 30 joules , then the final internal energy will be

Options

A. 2 j
B. `-18` j
C. `10j`
D. `58j`

Answer

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

\[ \Delta U = Q + W \]

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

### Step 1: Identify the values given in the problem
- Heat released by the gas, \(Q = -30 \, \text{J}\) (since the gas releases heat, it's negative)
- Work done on the gas, \(W = -10 \, \text{J}\) (since work is done on the gas, it's negative)
- Initial internal energy, \(U_i = 30 \, \text{J}\)

### Step 2: Calculate the change in internal energy
Using the first law of thermodynamics:

\[ \Delta U = Q + W \]

Substituting the values:

\[ \Delta U = (-30 \, \text{J}) + (-10 \, \text{J}) = -40 \, \text{J} \]

### Step 3: Calculate the final internal energy
The change in internal energy is related to the initial and final internal energy by the equation:

\[ \Delta U = U_f - U_i \]

Rearranging gives us:

\[ U_f = U_i + \Delta U \]

Substituting the known values:

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

### Final Answer
The final internal energy of the gas is:

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