Question

A balloon is filled at 27$^\circ$C and 1 atmospheric pressure by volume 500 m$^3$ helium gas. At -3$^\circ$C and 0.5 atmospheric pressure, the volume of helium gas will be ______.

• A. 500 m$^3$
• B. 700 m$^3$
• C. 900 m$^3$
• D. 1000 m$^3$

Hint:

The most common mistake in gas law problems is forgetting to convert Celsius to Kelvin. Remember, $0^\circ\text{C}$ would cause division by zero if left unconverted! Always use absolute temperature.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We are dealing with a fixed mass of an ideal gas undergoing changes in pressure, volume, and temperature. We must use the combined gas law to find the new volume.

Step 2: Key Formula or Approach:
The Combined Gas Law states that for a fixed amount of gas:
$$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$$
Temperatures must strictly be converted to the absolute Kelvin scale ($K = ^\circ\text{C} + 273$).

Step 3: Detailed Explanation:
1. Identify the initial state ($1$):
$P_1 = 1 \text{ atm}$
$V_1 = 500 \text{ m}^3$
$T_1 = 27^\circ\text{C} = 27 + 273 = 300 \text{ K}$
2. Identify the final state ($2$):
$P_2 = 0.5 \text{ atm}$
$T_2 = -3^\circ\text{C} = -3 + 273 = 270 \text{ K}$
$V_2 = ?$
3. Substitute the values into the combined gas law:
$$\frac{1 \times 500}{300} = \frac{0.5 \times V_2}{270}$$
Simplify the left side:
$$\frac{5}{3} = \frac{0.5 V_2}{270}$$
Rearrange to solve for $V_2$:
$$V_2 = \frac{5}{3} \times \frac{270}{0.5}$$
$$V_2 = \frac{5 \times 90}{0.5} = \frac{450}{0.5}$$
Dividing by 0.5 is the same as multiplying by 2:
$$V_2 = 450 \times 2 = 900 \text{ m}^3$$

Step 4: Final Answer:
The new volume will be 900 m$^3$, matching option (c).