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geniusboy [140]
3 years ago
10

An aerosol can contains 800.0 mL of compressed gas at 10.2 atm pressure. When the gas is sprayed into a large plastic bag, the b

ag inflates to a volume of 4.14 L. What is the pressure of gas (atm) inside the plastic bag if the temperature did not change and all of the gas was transferred completely?
Chemistry
2 answers:
Roman55 [17]3 years ago
5 0

Answer:

The pressure of the gas inside the plastic bag is 1.971 atm

Explanation:

Since temperature did not change, it means it is constant. This will then be Boyle's law. Thus

P1V1 =P2V2

Where P1 is the initial pressure = 10.2 ATM, P2 is the final pressure which is to be calculated.

V1 is the initial volume = 800.0 mL and V2 is the final volume of the gas = 4.14 L

From the formula,

P2 = P1V1/V2

= 10.2×800÷4.14

= 1.971 atm

Klio2033 [76]3 years ago
5 0

Answer:

Pressure of gas in the plastic bag is 1.97 atm

Explanation:

It is possible to answer this question by using Boyle's law: Boyle's law describes that, for ideal gases, the pressure of the gas is inversely proportional to the volume it occupies. Is expressed as:

P1V1 = P2V2

Where:

P1 is the initial pressure of the gas (In the problem, 10.2 atm).

V1 is initial volume (800.0mL)

V2 is final volume (4.14L ≡ 4140mL)

And P2 is the final pressure.

Replacing:

10.2 atm×800.0mL = 4140mL×P2

<em>P2 = 1.97 atm</em>

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The slow step reaction is given as:

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Now, the expression for the rate of reaction of fast reaction is:

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Substitute the value of [NOBr_{2}] in equation (1), we get:

\frac{d(NOBr)}{dt}=k_{2}[NOBr_{2}][NO]

=k_{2} \frac{k_{1}}{k_{-1}}[NO][Br_{2}][NO]

= \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}]

Thus, rate law of formation of NOBr in terms of reactants is given by \frac{k_{1}k_{2}}{k_{-1}}[NO]^{2}[Br_{2}].









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