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IRISSAK [1]
2 years ago
10

A 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, what is the volume? *

Chemistry
1 answer:
elena-14-01-66 [18.8K]2 years ago
3 0

The new volume when pressure increases to 2,030 kPa is 0.8L

BOYLE'S LAW:

The new volume of a gas can be calculated using Boyle's law equation:

P1V1 = P2V2

Where;

  1. P1 = initial pressure (kPa)
  2. P2 = final pressure (kPa)
  3. V1 = initial volume (L)
  4. V2 = final volume (L)

According to this question, a 4.0 L balloon has a pressure of 406 kPa. When the pressure increases to 2,030 kPa, the volume is calculated as:

406 × 4 = 2030 × V2

1624 = 2030V2

V2 = 1624 ÷ 2030

V2 = 0.8L

Therefore, the new volume when pressure increases to 2,030 kPa is 0.8L.

Learn more about Boyle's law calculations at: brainly.com/question/1437490?referrer=searchResults

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Answer: 72.4 kJ/mol

Explanation:

The balanced chemical reaction is,

C_3H_8(g)+5O_2(g)\rightarrow 3CO_2(g)+4H_2O(l)  \Delta H=-2220.1kJ/mol

The expression for enthalpy change is,

\Delta H=\sum [n\times \Delta H_f(product)]-\sum [n\times \Delta H_f(reactant)]

\Delta H=[(n_{CO_2}\times \Delta H_{CO_2})+(n_{H_2O}\times \Delta H_{H_2O})]-[(n_{O_2}\times \Delta H_{O_2})+(n_{C_3H_8}\times \Delta H_{C_3H_8})]

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Now put all the given values in this expression, we get

-2220.1=[(3\times -393.5)+(4\times -241.8)]-[(5\times 0)+(1\times \Delta H_{C_3H_8})]

\Delta H_{C_3H_8}=72.4kJ/mol

Therefore, the heat of formation of propane is 72.4 kJ/mol

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