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A balloon that had a volume of [tex]$3.50 \, L$[/tex] at [tex]$25.0^{\circ} C$[/tex] is placed in a hot room at [tex][tex]$40.0^{\circ} C$[/tex][/tex]. If the pressure remains constant at [tex]$1.00 \, atm$[/tex], what is the new volume of the balloon in the hot room?

Use [tex]\frac{V_1}{T_1} = \frac{V_2}{T_2}[/tex].

A. [tex]2.19 \, L[/tex]
B. [tex]3.33 \, L[/tex]
C. [tex]3.68 \, L[/tex]
D. [tex]5.60 \, L[/tex]



Answer :

GinnyAnswer
To find the new volume of the balloon when placed in a hot room, we can use Charles's Law, which relates the volume and temperature of a gas at constant pressure. The law is given by:

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

where:
- [tex]\( V_1 \)[/tex] is the initial volume of the gas,
- [tex]\( T_1 \)[/tex] is the initial temperature of the gas (in Kelvin),
- [tex]\( V_2 \)[/tex] is the final volume of the gas,
- [tex]\( T_2 \)[/tex] is the final temperature of the gas (in Kelvin).

First, let's convert the given temperatures from Celsius to Kelvin:
[tex]\[ T_1 = 25.0^{\circ}C + 273.15 = 298.15\, K \][/tex]
[tex]\[ T_2 = 40.0^{\circ}C + 273.15 = 313.15\, K \][/tex]

Next, we rearrange the formula to solve for the final volume [tex]\( V_2 \)[/tex]:
[tex]\[ V_2 = V_1 \times \frac{T_2}{T_1} \][/tex]

Substitute the known values into the equation:
[tex]\[ V_2 = 3.50\, L \times \frac{313.15\, K}{298.15\, K} \][/tex]

Now, we calculate:
[tex]\[ V_2 \approx 3.50 \times 1.050351 \][/tex]
[tex]\[ V_2 \approx 3.676 \][/tex]

Hence, the new volume of the balloon in the hot room is approximately [tex]\( 3.68 \, L \)[/tex].

From the given options:
- 2.19 L
- 3.33 L
- 3.68 L
- 5.60 L

The correct answer is:
[tex]\[ \boxed{3.68} \][/tex]

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