Question

A balloon filled with helium has a volume of 4.5 × 103 L at 25°C. What volume will the balloon occupy at 50°C if the pressure surrounding the balloon remains constant?

Answer 1

Answer:

$V_2 = 4.87 * 10^3$

Explanation:

This question is an illustration of ideal Gas Law;

The given parameters are as follows;

Initial Temperature = 25C

Initial Volume = 4.5 * 10³L

Required

Calculate the volume when temperature is 50C

NB: Pressure remains constant;

Ideal Gas Law states that;

$PV = nRT$

The question states that the pressure is constant; this implies that the constant in the above formula are P, R and n

Divide both sides by PT

$\frac{PV}{PT} = \frac{nRT}{PT}$

$\frac{V}{T} = \frac{nR}{P}$

Represent $\frac{nR}{P}$ with k

$\frac{V}{T} = k$

$k = \frac{V_1}{T_1} = \frac{V_2}{T_2}$

At this point, we can solve for the required parameter using the following;

$\frac{V_1}{T_1} = \frac{V_2}{T_2}$

Where V1 and V2 represent the initial & final volume and T1 and T2 represent the initial and final temperature;

From the given parameters;

V1 = 4.5 * 10³L

T1 = 25C

T2 = 50C

Convert temperatures to degree kelvin

V1 = 4.5 * 10³L

T1 = 25 +273 = 298K

T2 = 50 + 273 = 323K

Substitute values for V1, T1 and T2 in $\frac{V_1}{T_1} = \frac{V_2}{T_2}$

$\frac{4.5 * 10^3}{298} = \frac{V_2}{323}$

Multiply both sides by 323

$323 * \frac{4.5 * 10^3}{298} = \frac{V_2}{323} * 323$

$323 * \frac{4.5 * 10^3}{298} = V_2$

$V_2 = 323 * \frac{4.5 * 10^3}{298}$

$V_2 = \frac{323 * 4.5 * 10^3}{298}$

$V_2 = \frac{1453.5 * 10^3}{298}$

$V_2 = 4.87 * 10^3$

Hence, the final volume at 50C is $V_2 = 4.87 * 10^3$