Five moles of an ideal monoatomic gas with an initial temperature of `150^@C` expand and in the process absorb 1500 J of heat and does 2500 J of work. The final temperature of the gas in `.^@C` is (ideal gas constant `R = 8.314 J K^(-1)mol^(-1))`

Options

A. 134
B. 126
C. 144
D. 166

Answer

To solve the problem, we will use the first law of thermodynamics and the relationship between internal energy change and temperature change for an ideal gas. Here are the steps to find the final temperature of the gas:

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states:
\[ \Delta Q = \Delta U + \Delta W \]
Where:
- \(\Delta Q\) is the heat absorbed by the system.
- \(\Delta U\) is the change in internal energy.
- \(\Delta W\) is the work done by the system.

### Step 2: Substitute the Given Values
From the problem, we have:
- \(\Delta Q = 1500 \, J\)
- \(\Delta W = 2500 \, J\)

Substituting these values into the first law equation:
\[ 1500 = \Delta U + 2500 \]

### Step 3: Solve for Change in Internal Energy (\(\Delta U\))
Rearranging the equation gives:
\[ \Delta U = 1500 - 2500 = -1000 \, J \]

### Step 4: Relate Change in Internal Energy to Temperature Change
For an ideal monoatomic gas, the change in internal energy is given by:
\[ \Delta U = n C_V \Delta T \]
Where:
- \(n\) is the number of moles (5 moles).
- \(C_V\) for a monoatomic ideal gas is \(\frac{3}{2} R\).
- \(R = 8.314 \, J \, K^{-1} \, mol^{-1}\).

Calculating \(C_V\):
\[ C_V = \frac{3}{2} \times 8.314 = 12.471 \, J \, K^{-1} \, mol^{-1} \]

### Step 5: Substitute Values into the Internal Energy Equation
Substituting the values into the equation for \(\Delta U\):
\[ -1000 = 5 \times 12.471 \times \Delta T \]

### Step 6: Solve for \(\Delta T\)
Rearranging gives:
\[ \Delta T = \frac{-1000}{5 \times 12.471} \]
Calculating the denominator:
\[ 5 \times 12.471 = 62.355 \]
Now substituting back:
\[ \Delta T = \frac{-1000}{62.355} \approx -16.03 \, K \]

### Step 7: Calculate the Final Temperature
We know that:
\[ \Delta T = T_{final} - T_{initial} \]
Given \(T_{initial} = 150 \, ^\circ C\):
\[ -16.03 = T_{final} - 150 \]
Rearranging gives:
\[ T_{final} = 150 - 16.03 \approx 133.97 \, ^\circ C \]

### Final Answer
The final temperature of the gas is approximately:
\[ T_{final} \approx 134 \, ^\circ C \]