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 properties of an ideal monoatomic gas. Let's break it down step by step.

### Step 1: Understand the First Law of Thermodynamics
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):
\[ \Delta U = Q - W \]

### Step 2: Identify Given Values
We are given:
- Number of moles (n) = 5 moles
- Initial temperature (T_initial) = 150°C = 150 + 273.15 = 423.15 K
- Heat absorbed (Q) = 1500 J
- Work done (W) = 2500 J
- Ideal gas constant (R) = 8.314 J/(K·mol)

### Step 3: Calculate Change in Internal Energy
Using the first law of thermodynamics:
\[ \Delta U = Q - W = 1500 J - 2500 J = -1000 J \]

### Step 4: Relate Change in Internal Energy to Temperature Change
For a monoatomic ideal gas, the change in internal energy is given by:
\[ \Delta U = \frac{3}{2} n R \Delta T \]
where ΔT is the change in temperature.

### Step 5: Substitute Values into the Equation
Substituting the values we have:
\[ -1000 J = \frac{3}{2} \times 5 \times 8.314 \times \Delta T \]
Calculating the right side:
\[ -1000 J = \frac{15}{2} \times 8.314 \times \Delta T \]
\[ -1000 J = 62.355 \Delta T \]

### Step 6: Solve for ΔT
Now, we can solve for ΔT:
\[ \Delta T = \frac{-1000 J}{62.355} \approx -16.03 K \]

### Step 7: Calculate Final Temperature
The final temperature (T_final) can be calculated as:
\[ T_{final} = T_{initial} + \Delta T = 423.15 K - 16.03 K \approx 407.12 K \]
Converting back to Celsius:
\[ T_{final} = 407.12 K - 273.15 \approx 134°C \]

### Final Answer
The final temperature of the gas is approximately:
\[ \boxed{134°C} \]