Question

Unless otherwise instructed, you may use the periodic table in the Chemistry: Problems and Solutions book for this question. You also may use the formulas on pages 238–243.

A student increases the temperature of a 556 cm3 balloon from 278 K to 308 K.

Assuming constant pressure, what should the new volume of the balloon be?

Responses

320 cm3

320 cm 3

501 cm3

501 cm 3

616 cm3

616 cm 3

1232 cm3

Answers

GPT-4o mini
To find the new volume of the balloon when the temperature is increased at constant pressure, we can use Charles's Law, which states that the volume of a gas is directly proportional to its temperature (in Kelvin) when the pressure is constant. The formula can be expressed as:

\[
\frac{V_1}{T_1} = \frac{V_2}{T_2}
\]

Where:
- \( V_1 \) is the initial volume
- \( T_1 \) is the initial temperature
- \( V_2 \) is the final volume
- \( T_2 \) is the final temperature

Given:
- \( V_1 = 556 \, \text{cm}^3 \)
- \( T_1 = 278 \, \text{K} \)
- \( T_2 = 308 \, \text{K} \)

We need to find \( V_2 \).

Rearranging the formula to solve for \( V_2 \):

\[
V_2 = V_1 \times \frac{T_2}{T_1}
\]

Now, substituting the known values:

\[
V_2 = 556 \, \text{cm}^3 \times \frac{308 \, \text{K}}{278 \, \text{K}}
\]

Calculating \( \frac{308}{278} \):

\[
\frac{308}{278} \approx 1.108
\]

Now calculate \( V_2 \):

\[
V_2 \approx 556 \, \text{cm}^3 \times 1.108 \approx 617.64 \, \text{cm}^3
\]

Rounding it off, \( V_2 \) is approximately 618 cm³.

Since this value does not perfectly match any of the provided responses, the closest option among the choices would be **616 cm³**.

Therefore, the new volume of the balloon should be **616 cm³**.