At `0^(@)C` and one atm pressure, a gas occupies 100 cc. If the pressure is increased to one and a half-time and temprature is increased by one-third of absolute temperature, then final volume of the gas will be:

Options

A. 80 cc
B. 88.9 cc
C. 66.7 cc
D. 100 cc

Explanation

To solve the problem, we will use the ideal gas law, which states that:

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

Where:
- \( P_1 \) = initial pressure
- \( V_1 \) = initial volume
- \( T_1 \) = initial temperature
- \( P_2 \) = final pressure
- \( V_2 \) = final volume
- \( T_2 \) = final temperature

### Step 1: Identify the initial conditions
- Initial volume \( V_1 = 100 \, \text{cc} \)
- Initial temperature \( T_1 = 0^\circ C = 273 \, \text{K} \)
- Initial pressure \( P_1 = 1 \, \text{atm} \)

### Step 2: Determine the final conditions
- The pressure is increased to one and a half times the initial pressure:
\[ P_2 = 1.5 \times P_1 = 1.5 \times 1 \, \text{atm} = 1.5 \, \text{atm} \]

- The temperature is increased by one-third of the absolute temperature:
\[ T_2 = T_1 + \frac{1}{3} T_1 = T_1 \left(1 + \frac{1}{3}\right) = \frac{4}{3} T_1 = \frac{4}{3} \times 273 \, \text{K} = 364 \, \text{K} \]

### Step 3: Substitute the values into the ideal gas law equation
Using the ideal gas law:
\[ \frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2} \]
Substituting the known values:
\[ \frac{1 \, \text{atm} \times 100 \, \text{cc}}{273 \, \text{K}} = \frac{1.5 \, \text{atm} \times V_2}{364 \, \text{K}} \]

### Step 4: Rearranging to solve for \( V_2 \)
Cross-multiplying gives:
\[ 1 \times 100 \times 364 = 1.5 \times V_2 \times 273 \]
\[ 36400 = 1.5 \times 273 \times V_2 \]
Now, calculate \( 1.5 \times 273 \):
\[ 1.5 \times 273 = 409.5 \]
So we have:
\[ 36400 = 409.5 \times V_2 \]

### Step 5: Solve for \( V_2 \)
\[ V_2 = \frac{36400}{409.5} \approx 88.9 \, \text{cc} \]

### Final Answer
The final volume of the gas is approximately \( V_2 \approx 88.9 \, \text{cc} \).

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