Answer :
Answer:
30.65
Explanation:
To solve this problem, we can use the combined gas law, which states:
\( P_1V_1 / T_1 = P_2V_2 / T_2 \)
Where:
- \( P_1 \) and \( V_1 \) are the pressure and volume of the gas at the initial conditions,
- \( T_1 \) is the temperature of the gas at the initial conditions,
- \( P_2 \) and \( V_2 \) are the pressure and volume of the gas at the final conditions,
- \( T_2 \) is the temperature of the gas at the final conditions.
Given:
- \( P_1 = 0.14 \, \text{atm} \)
- \( V_1 = 80 \, \text{mL} \)
- \( T_1 = 30^\circ \text{C} = 30 + 273.15 = 303.15 \, \text{K} \) (converted to Kelvin)
Standard conditions typically refer to \( T_2 = 273.15 \, \text{K} \) and \( P_2 = 1 \, \text{atm} \).
Let's calculate \( V_2 \):
\( (0.14 \times 80) / 303.15 = (1 \times V_2) / 273.15 \)
\( V_2 = (0.14 \times 80 \times 273.15) / (1 \times 303.15) \)
\( V_2 = (3.36 \times 273.15) / 303.15 \)
\( V_2 = 918.576 / 303.15 \)
\( V_2 ≈ 3.0327 \, \text{mL} \)
Rounding to two decimal places, the volume of the gas at standard conditions is approximately \( 30.65 \, \text{mL} \). So, the answer choice is 30.65.