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Aleks04 [339]
3 years ago
7

Determine which law is appropriate for solving the following problem. What temperature will 215 mL of a gas at 20°C and 1 atm pr

essure attain when it is subject to 1.5 atm of pressure? what Law is used: Combined gas law, Charles Law, or boyles law?
Chemistry
2 answers:
Murljashka [212]3 years ago
5 0

Answer : The temperature will be 439.5 K and combined gas law is used.

Solution : Given,

Volume = 215 ml

Initial conditions,

T_1=20^oC=273+20=293K   (0^oC=273K)

P_1=1atm

P_2=1.5atm

Using combined gas law,

\frac{PV}{T}=constant

where,

P = pressure of gas

V = volume of gas

T = temperature of gas

As per the problem, at constant volume the combined gas law is expressed as,

\frac{P_1}{T_1}=\frac{P_2}{T_2}

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

\frac{1atm}{293K}=\frac{1.5atm}{T_2}

By rearranging the term, we get the value of T_2.

T_2=439.5K

Therefore, the temperature will be 439.5 K and combined gas law is used.

loris [4]3 years ago
4 0

Answer:

The Combined Gas Law

Explanation:

Let’s compare the two conditions.

p₁ = 1     atm; V₁ = 215 mL; T₁ = 20 °C

p₂ = 1.5 atm; V₂ = 215 mL; T₂= ?

The only variables are pressure and temperature.

The most appropriate law would be <em>Gay-Lussac’s Law</em> (p₁/T₁ = p₂/T₂), but that’s not an option.

The only other option involving p and T is the <em>Combined Gas Law</em>:

p₁V₁/T₁ = p₂V₂/T₂

Since V₂ = V₁, we can write

p₁V₁/T₁ = p₂V₁/T₂     Divide both sides by V₁

    p₁/T₁= p₂/T₂         Multiply both sides by T₂

p₁T₂/T₁ = p₂               Multiply both sides by T₁

    p₁T₂= p₂T₁            Divide both sides by p₁

      T₂ = T₁ × p₂/p₁

Now, convert your temperature to kelvins, insert the values into the formula, and calculate the new temperature.

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HELPPPPP PLEASE
andreyandreev [35.5K]

Answer:

V=1.18 L

Explanation:

Alternative solution:

Calculate how many moles

2.34

g

C

O

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is

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g

⋅

1

m

o

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44

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=

0.053

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Simple mole ratios tell us that

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0.053

m

o

l

, we times

22.4

L

m

o

l

by

0.053

m

o

l

.

22.4

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m

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⋅

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Answer link

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