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3 years ago

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The volume of a gas is 450 mL when its pressure is 1.00 atm. If the temperature of the gas does not change, what is the pressure

when its volume is changed to 2.00 L?
Use: P1V1=P2V2

A) 0.225 atm
B) 0.444 atm
C) 2.25 atm
D) 4.44 atm

1 answer:

juin [17]3 years ago

6 0

Answer:

A) 0.225atm

Explanation:

P1V1 = P2V2

V1 = 450ml

P1 = 1.0atm

V2 = 2L = 2 X 100 = 2000ml

P2 =?

1.0 X 450 = P2 X 2000

P2 = (1.0 X 450)/2000

= 0.225atm

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<u>Answer:</u> 6.57 L of solution can be made.

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The specific heats of monoatomic gases, measured in joules per mol-Kelvin, are represented by the following expressions:

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$c_{v} = \frac{3}{2}\cdot R_{u}$ (1)

Isobaric (Constant pressure)

$c_{p} = \frac{5}{2}\cdot R_{u}$ (2)

Where $R_{u}$ is the ideal gas constant, measured in pascal-cubic meters per mol-Kelvin.

Under the assumption of ideal gas, we notice the following relationships:

1) Temperature is directly proportional to pressure.

2) Temperature is directly proportional to volume.

Now we proceed to find all required temperatures below:

(i) <em>Alice heats the gas at constant volume until its pressure is doubled</em>:

$\frac{T_{2}}{T_{1}} = \frac{P_{2}}{P_{1}}$ (3)

( $\frac{P_{2}}{P_{1}} = 2$ , $T_{1} = 273.15\,K$ )

$T_{2} = \frac{P_{2}}{P_{1}} \times T_{1}$

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$T_{2} = 546.3\,K$

(ii) <em>Bob further heats the gas at constant pressure until its volume is doubled</em>:

$\frac{T_{3}}{T_{2}} =\frac{V_{3}}{V_{2}}$ (4)

( $\frac{V_{3}}{V_{2}} = 2$ , $T_{2} = 546.3\,K$ )

$T_{3} = \frac{V_{3}}{V_{2}}\times T_{2}$

$T_{3} = 2\times 546.3\,K$

$T_{3} = 1092.6\,K$

Finally, the heat added to the gas ( $Q$ ), measured in joules, for the entire process is:

$Q = n\cdot [c_{v}\cdot (T_{2}-T_{1})+c_{p}\cdot (T_{3}-T_{2})]$ (5)

If we know that $R_{u} = 8.314\,\frac{Pa\cdot m^{3}}{mol\cdot K}$ , $n = 2\,mol$ , $T_{1} = 273.15\,K$ , $T_{2} = 546.3\,K$ and $T_{3} = 1092.6\,K$ , the heat added to the gas for the entire process is:

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$Q = 29273.178\,J$

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