Question

A balloon inflated in a room at 25oC has a volume of 3.00 L. If the balloon is heated at a constant pressure, at what temperature will the volume of the balloon double?

1. 596 K
2. 323 K
3. 298 K
4. 149 K

Answer 1

596K

Step-by-step explanation:

Using Charles law equation;

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

V1 = 3.00 L

V2 = double of V1 = 2 × 3.00 = 6.00 L

T1 = 25°C = 25 + 273 = 298K

T2 = ?

Using V1/T1 = V2/T2

3/298 = 6/T2

Cross multiply

298 × 6 = 3 × T2

1788 = 3T2

T2 = 1788 ÷ 3

T2 = 596K