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 \( PV = nRT \). We will analyze the initial and final states of the gas and apply the formula accordingly.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- Initial Volume, \( V_1 = 100 \, \text{cc} \)
- Initial Pressure, \( P_1 = 1 \, \text{atm} \)
- Initial Temperature, \( T_1 = 0^\circ C = 273 \, \text{K} \) (Convert Celsius to Kelvin)

2. **Determine 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:
\[ T_2 = T_1 + \frac{1}{3} T_1 = T_1 \left(1 + \frac{1}{3}\right) = T_1 \times \frac{4}{3} = 273 \times \frac{4}{3} = 364 \, \text{K} \]

3. **Apply the Ideal Gas Law**:
- Using the relation \( P_1 V_1 / T_1 = P_2 V_2 / T_2 \), we can rearrange it to find \( V_2 \):
\[ V_2 = \frac{P_1 V_1 T_2}{P_2 T_1} \]

4. **Substituting the Values**:
- Substitute \( P_1 = 1 \, \text{atm} \), \( V_1 = 100 \, \text{cc} \), \( T_2 = 364 \, \text{K} \), \( P_2 = 1.5 \, \text{atm} \), and \( T_1 = 273 \, \text{K} \):
\[ V_2 = \frac{1 \times 100 \times 364}{1.5 \times 273} \]

5. **Calculate \( V_2 \)**:
- Calculate the numerator:
\[ 1 \times 100 \times 364 = 36400 \]
- Calculate the denominator:
\[ 1.5 \times 273 = 409.5 \]
- Now, divide the numerator by the denominator:
\[ V_2 = \frac{36400}{409.5} \approx 88.9 \, \text{cc} \]

### Final Answer:
The final volume of the gas, \( V_2 \), is approximately **88.9 cc**.