A gas at a pressure of 5.0 atm is heated from `0^(@) " to" 546^(@)C` and is simultaneously compressed to one-third of its original volume. Hence final pressure is :

Options

A. 15.0 atm
B. 30.0 atm
C. 45.0 atm
D. `5//9` atm

Explanation

To solve the problem, we will use the combined gas law, which relates the pressure, volume, and temperature of a gas. The equation is given by:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

### Step 1: Identify the initial conditions
- Initial pressure, \( P_1 = 5.0 \, \text{atm} \)
- Initial temperature, \( T_1 = 0^\circ C = 273 \, \text{K} \)
- Initial volume, \( V_1 \) (we will keep it as \( V_1 \) since we don't need its exact value)

### Step 2: Identify the final conditions
- The gas is heated to \( T_2 = 546^\circ C = 546 + 273 = 819 \, \text{K} \)
- The volume is compressed to one-third of its original volume, so \( V_2 = \frac{1}{3} V_1 \)

### Step 3: Substitute the known values into the equation
Using the combined gas law:

\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]

Substituting the known values:

\[ \frac{5.0 \, \text{atm} \cdot V_1}{273 \, \text{K}} = \frac{P_2 \cdot \left(\frac{1}{3} V_1\right)}{819 \, \text{K}} \]

### Step 4: Simplify the equation
We can cancel \( V_1 \) from both sides:

\[ \frac{5.0 \, \text{atm}}{273} = \frac{P_2 \cdot \frac{1}{3}}{819} \]

### Step 5: Solve for \( P_2 \)
Rearranging the equation to solve for \( P_2 \):

\[ P_2 = \frac{5.0 \, \text{atm} \cdot 819}{273} \cdot 3 \]

### Step 6: Calculate \( P_2 \)
Calculating the right-hand side:

\[ P_2 = \frac{5.0 \cdot 819 \cdot 3}{273} \]

Calculating \( 5.0 \cdot 819 = 4095 \):

\[ P_2 = \frac{4095 \cdot 3}{273} = \frac{12285}{273} \approx 45 \, \text{atm} \]

### Final Answer
Thus, the final pressure \( P_2 \) is approximately **45 atm**.

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