At `27^(@)C` and one bar pressure a has volume `VL`. What will be its volume at `177 ^(2)C` and under pressure of `1.5"bar"`?

Explanation

To solve the problem, we will use the Ideal Gas Law, which states that for a given amount of gas, the relationship between pressure, volume, and temperature is given by the equation:

\[ P_1 V_1 / T_1 = P_2 V_2 / T_2 \]

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature (in Kelvin)
- \( P_2 \) = final pressure
- \( V_2 \) = final volume (what we need to find)
- \( T_2 \) = final temperature (in Kelvin)

### Step 1: Convert temperatures from Celsius to Kelvin
1. **Initial Temperature (T1)**:
\[ T_1 = 27^\circ C + 273 = 300 \, K \]

2. **Final Temperature (T2)**:
\[ T_2 = 177^\circ C + 273 = 450 \, K \]

### Step 2: Identify the given values
- Initial volume \( V_1 = V \) liters
- Initial pressure \( P_1 = 1 \, \text{bar} \)
- Final pressure \( P_2 = 1.5 \, \text{bar} \)
- Initial temperature \( T_1 = 300 \, K \)
- Final temperature \( T_2 = 450 \, K \)

### Step 3: Rearrange the Ideal Gas Law to solve for \( V_2 \)
Using the Ideal Gas Law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Rearranging gives:
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the equation:
\[ V_2 = \frac{(1 \, \text{bar}) \cdot (V \, \text{liters}) \cdot (450 \, K)}{(1.5 \, \text{bar}) \cdot (300 \, K)} \]

### Step 5: Simplify the expression
\[ V_2 = \frac{1 \cdot V \cdot 450}{1.5 \cdot 300} \]
\[ V_2 = \frac{450V}{450} = \frac{V}{1.5} \]
\[ V_2 = \frac{2}{3} V \]

### Final Answer:
The volume \( V_2 \) at \( 177^\circ C \) and \( 1.5 \, \text{bar} \) will be \( \frac{2}{3} V \) liters.

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