Question

A balloon contains \(1500\ \mathrm{m}^3\) of helium at \(27^{\circ}\mathrm{C}\) and 4 atmospheric pressure. The volume of helium at \(-3^{\circ}\mathrm{C}\) temperature and 2 atmospheric pressure will be

• A. \(2700\ \mathrm{m}^3\)
• B. \(1900\ \mathrm{m}^3\)
• C. \(1700\ \mathrm{m}^3\)
• D. \(1500\ \mathrm{m}^3\)

Hint:

Always convert temperature to Kelvin: \(T(K) = T(^{\circ}C) + 273\).

Answer 1

The Correct Option is A

To find the volume of helium at a different temperature and pressure, we will use the combined gas law, which relates the pressure, volume, and temperature of a gas. The formula for the combined gas law is:

\(\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}\)

1. Given:
   • Initial volume, \(V_1 = 1500\ \mathrm{m}^3\)
   • Initial pressure, \(P_1 = 4\ \text{atm}\)
   • Initial temperature, \(T_1 = 27^{\circ}\text{C} = 300\ \text{K}\) (Convert Celsius to Kelvin by adding 273)
   • Final pressure, \(P_2 = 2\ \text{atm}\)
   • Final temperature, \(T_2 = -3^{\circ}\text{C} = 270\ \text{K}\)
2. We need to find the final volume, \(V_2\).

Rearrange the combined gas law formula to solve for \(V_2\):

\(V_2 = \frac{P_1 V_1 T_2}{P_2 T_1}\)

1. Substitute the given values into the equation:

\(V_2 = \frac{4 \times 1500 \times 270}{2 \times 300}\)

1. Simplify the expression:

\(V_2 = \frac{1620000}{600} = 2700\ \mathrm{m}^3\)

1. The volume of helium at \(-3^{\circ}\mathrm{C}\) and 2 atmospheric pressure is \(2700\ \mathrm{m}^3\).

Therefore, the correct answer is \(2700\ \mathrm{m}^3\).