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

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

Accepted Answer

{ "quality": "clear", "question_count": 1, "skills_matched": [ "step_cot", "multiple_choice_formatter", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Charless Law" ], "new_concepts": [], "current_concepts": [ "Charless Law", "Gas Laws" ] } ## 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 $$ \begin{aligned} V_2 &= V_1 \cdot \frac{T_2}{T_1} \ V_2 &= 3.50\text{ L} \cdot \frac{313.15\text{ K}}{298.15\text{ K}} \ V_2 &\approx 3.68\text{ L} \end{aligned} $$

2.19 L

3.33 L

3.68 L

5.60 L

{ "subject": "Natural Science", "question_type": "Multiple Choice", "knowledge_point": [ "Natural Science", "Chemistry", "Charless Law" ] }

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Question

Explanation:

Convert temperatures to Kelvin

Solve for the final volume

Answer: