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Artyom0805 [142]
3 years ago
15

To what temperature in kelvin will a 2.3 L balloon have to be heated to expand to a volume of 400 L Assume the initial temperatu

re of the balloon is 25 degrees celsius.
Chemistry
1 answer:
olchik [2.2K]3 years ago
7 0

Answer:

T₂ = 51826.1 K

Explanation:

Given data:

Initial Volume = 2.3 L

Final volume = 400 L

Initial temperature = 25 °C (25+ 273 = 298 K)

Final temperature = ?

Solution:

V₁/T₁ = V₂/T₂

T₂ = V₂ T₁/V₁

T₂ =  400 L . 298 K / 2.3 L

T₂ = 119200 K. L / 2.3 L

T₂ = 51826.1 K

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A sample of neon occupies a volume of 752 ml at 25 0
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V_{initial} = 752\:mL
T_{initial} = 25.0^0C
converting to Kelvin
TK = TC + 273
TK = 25.0 + 273 → TK = 298.0 → T_{initial} = 298.0\:K
V_{final} = ? (in\:milliliters)
T_{final} = 50.0^0C
TK = TC + 273
TK = 50.0 + 273 → TK = 323.0 → T_{final} = 323.0\:K

By the first Law of Charles and Gay-Lussac, we have: 
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }

Solving:
\frac{ V_{i} }{ T_{i} } = \frac{ V_{f} }{ T_{f} }
\frac{ 752 }{ 298.0 } = \frac{ V_{f} }{ 323.0 }
Product of extremes equals product of means:
298.0* V_{f} = 752*323.0
298.0 V_{f} = 242896
V_{f} = \frac{242896}{298.0}
\boxed{\boxed{V_{f} \approx 815.08\:mL}}\end{array}}\qquad\quad\checkmark
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