Question

a balloon that had a volume of 3.50 l at 25.0°c is placed in a hot room at 40.0°c. if the pressure remains constant at 1.00 atm, what is the new volume of the balloon in the hot room?
use \\(\frac{v_1}{t_1} = \frac{v_2}{t_2}\\).

2.19 l
3.33 l
3.68 l
5.60 l

Explanation:

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<reasoning>

Convert temperatures to Kelvin

\[

$$\begin{aligned} T_1 &= 25.0^\circ\text{C} + 273.15 = 298.15\text{ K} \\ T_2 &= 40.0^\circ\text{C} + 273.15 = 313.15\text{ K} \end{aligned}$$

\]

Solve for the final volume

\[

$$\begin{aligned} V_2 &= V_1 \cdot \frac{T_2}{T_1} \\ V_2 &= 3.50\text{ L} \cdot \frac{313.15\text{ K}}{298.15\text{ K}} \\ V_2 &\approx 3.68\text{ L} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>2.19 L</mcq-option>
<mcq-option>3.33 L</mcq-option>
<mcq-correct>3.68 L</mcq-correct>
<mcq-option>5.60 L</mcq-option>
</answer>

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Answer:

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<reasoning>

Convert temperatures to Kelvin

\[

$$\begin{aligned} T_1 &= 25.0^\circ\text{C} + 273.15 = 298.15\text{ K} \\ T_2 &= 40.0^\circ\text{C} + 273.15 = 313.15\text{ K} \end{aligned}$$

\]

Solve for the final volume

\[

$$\begin{aligned} V_2 &= V_1 \cdot \frac{T_2}{T_1} \\ V_2 &= 3.50\text{ L} \cdot \frac{313.15\text{ K}}{298.15\text{ K}} \\ V_2 &\approx 3.68\text{ L} \end{aligned}$$

\]
</reasoning>

<answer>
<mcq-option>2.19 L</mcq-option>
<mcq-option>3.33 L</mcq-option>
<mcq-correct>3.68 L</mcq-correct>
<mcq-option>5.60 L</mcq-option>
</answer>

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