The pressure of given mass of a gas in a thermodynamic system is changed in such a way that 20 joule of heat is released from the gas and 8 joule of work is done on the gas. If the initial internal energy of the gas was 30 joule then final internal energy will be

Options

A. 2 joule
B. 42 joule
C. 18 joule
D. 58 joule

Answer

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

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

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

### Step 1: Identify the given values
- Heat released from the gas: \(\Delta Q = -20 \, \text{J}\) (negative because heat is released)
- Work done on the gas: \(\Delta W = -8 \, \text{J}\) (negative because work is done on the gas)
- Initial internal energy: \(U_i = 30 \, \text{J}\)

### Step 2: Write the first law of thermodynamics
Using the first law of thermodynamics:

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

We can rearrange this to find the change in internal energy:

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

### Step 3: Substitute the known values
Substituting the values we have into the equation:

\[ \Delta U = -20 \, \text{J} - (-8 \, \text{J}) \]

### Step 4: Simplify the equation
Now simplify the equation:

\[ \Delta U = -20 \, \text{J} + 8 \, \text{J} = -12 \, \text{J} \]

### Step 5: Calculate the final internal energy
The change in internal energy can also be expressed in terms of initial and final internal energy:

\[ \Delta U = U_f - U_i \]

Substituting the values we have:

\[ -12 \, \text{J} = U_f - 30 \, \text{J} \]

Now, solve for \(U_f\):

\[ U_f = -12 \, \text{J} + 30 \, \text{J} = 18 \, \text{J} \]

### Final Answer
The final internal energy \(U_f\) is:

\[ U_f = 18 \, \text{J} \]

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