Question

A balloon filled with 47 mL of hydrogen gas at 276 K is placed in a freezer. What will be the new volume if the temperature of the balloon is raised to 456 K?

Answer 1

`\blue{\huge {\mathrm{CHARLES' \; LAW}}}`

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`{\underline{\huge \mathbb{Q} {\large \mathrm {UESTION : }}}}`

• A balloon filled with 47 mL of hydrogen gas at 276 K is placed in a freezer. What will be the new volume if the temperature of the balloon is raised to 456 K?

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`{\underline{\huge \mathbb{A} {\large \mathrm {NSWER : }}}}`

• The new volume of the balloon at a temperature of 456 K is 77.7 mL.

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`{\underline{\huge \mathbb{S}{\large \mathrm {OLUTION : }}}}`

We can use Charles' Law to solve this problem, which states that at constant pressure, the volume of a gas is directly proportional to its absolute temperature:

• 
  `\sf(V_1)/(T_1) = (V_2)/(T_2)`

where:

• 
  `\sf V_1` is the initial volume (47 mL),
• 
  `\sf T_1` is the initial temperature (276 K),
• 
  `\sf V_2` is the final volume (unknown), and
• 
  `\sf T_2` is the final temperature (456 K).

Solving for
`\sf V_2`, we get:

`\begin{aligned}\sf V_2& =\sf ((V_1 \cdot T_2))/(T_1) \\& =\sf ((47\: mL \cdot 456\: K))/(276\: K) \\ & = \boxed{\bold{\:77.7\: mL\:}} \end{aligned}`

Therefore, the new volume of the balloon at a temperature of 456 K is 77.7 mL.

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`- \large\sf\copyright \: \large\tt{AriesLaveau}\large\qquad\qquad\qquad\qquad\qquad\qquad\tt 04/02/2023`

Answer 2

The new volume of the balloon will be approximately 77.7 mL (3 s.f.).

Step-by-step explanation:

We can assume that the pressure remains constant, since the balloon is not being compressed or expanded, and is simply being exposed to a different temperature. Therefore, to solve this problem, we can use Charles's Law, which relates the volume and temperature of a gas.

Charles's Law

`\boxed{\sf (V_1)/(T_1)=(V_2)/(T_2)}`

where:

• V₁ = initial volume
• T₁ = initial temperature (in kelvin)
• V₂ = final volume
• T₂ = final temperature (in kelvin)

Given the balloon is initially filled with 47 mL of hydrogen gas at 276 K, and its temperature is raised to 456 K:

• V₁ = 47 mL
• T₁ = 276 K
• T₂ = 456 K

Substitute the given values into the formula:

`\implies \sf (47\;mL)/(276\;K)=(V_2)/(456\;K)`

Solve for V₂:

`\implies \sf (47\;mL)/(276\;K) \cdot 456\;K=(V_2)/(456\;K) \cdot 456\;K`

`\implies \sf V_2=(47\;mL\cdot 456\;K)/(276\;K)`

`\implies \sf V_2=77.6521739...\;mL`

`\implies \sf V_2=77.7\; mL\; (3\;s.f.)`

Therefore, the new volume of the balloon will be approximately 77.7 mL when the temperature is raised to 456 K.