A balloon contains 500m³ of helium at 27°C and 1 atmosphere pressure. The volume of the helium at – 3°C temperature and 0.5 atmosphere pressure will be

The correct option is 900 m³.

Step-by-Step Explanation:

To find the final volume of the helium balloon under the new temperature and pressure conditions, we can use the Combined Gas Law. This law relates the pressure, volume, and temperature of a given amount of gas before and after a change in conditions.

The formula for the Combined Gas Law is:
$$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$$
where:

• _P1 and _P2 are the initial and final pressures
• _V1 and _V2 are the initial and final volumes
• _T1 and _T2 are the initial and final absolute temperatures (measured in Kelvin)

1. Identify the given values from the problem:
Initial Volume (_V1) = 500 m³
Initial Temperature (_t1) = 27°C
Initial Pressure (_P1) = 1 atm
Final Temperature (_t2) = –3°C
Final Pressure (_P2) = 0.5 atm

2. Convert temperatures from Celsius to Kelvin:
Temperature must always be in Kelvin (K) when using gas law formulas:
$$T_1=27+273=300\text{K}$$
$$T_2=-3+273=270\text{K}$$

3. Rearrange the formula to solve for the final volume (_V2):
$$V_2=\frac{P_1\times V_1\times T_2}{P_2\times T_1}$$

4. Substitute the values into the equation:
$$V_2=\frac{1\times 500\times 270}{0.5\times 300}$$
Simplify the numerator and the denominator:
$$V_2=\frac{135000}{150}$$
$$V_2=900{\text{m}}^3$$

Therefore, the volume of the helium at the new conditions will be 900 m³.