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2 years ago

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What would be the original temperature of a gas that has a volume of 2.0 L and a pressure of 2.0 ATM and an unknown temperature

that the volume increased to 3.5 L in its pressure decreased to 1.0 ATM if the final temperature is measured to be 11°C

1 answer:

eduard2 years ago

8 0

Answer:

= 51.57 °C

Explanation:

The combined gas equation shows that P₁V₁/T₁=P₂V₂/T₂ where P represents pressure, V represents volume and T absolute temperature

From the information provided in the question,

Since we are finding the initial or original temperature, we can make T₁ the subject of the formula.

T₁=P₁V₁T₂/P₂V₂

P₁=2.0 ATM

P₂= 1.0 ATM

V₁=2.0 L

V₂= 3.5 L

T₁=?

T₂= (11+273) K=284 K

Using these values in the formula:

T₁= (2.0 ATM × 2.0 L× 284 K)/(1.0 ATM × 3.5 L)

=324.57 K

324.57-273= 51.57 °C

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Formula used :

$Q=m\times c\times \Delta T$

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$Q=m\times c\times (T_2-T_1)$

where,

Q = heat released = ?

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$C_p$ = specific heat capacity of water = $4.184J/g^oC$

$T_1$ = initial temperature = $25.0^oC$

$T_2$ = final temperature = $29.83^oC$

Now put all the given value in the above formula, we get:

$Q=93.1g\times 4.184J/g^oC\times (29.83-25.00)^ioC$

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$\text{Moles of }Ba(OH)_2=\text{Concentration of }Ba(OH)_2\times \text{Volume of solution}$

$\text{Moles of }Ba(OH)_2=0.266M\times 0.0365L=9.71\times 10^{-3}mol$

and,

$\text{Moles of }HCl=\text{Concentration of }HCl\times \text{Volume of solution}$

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The balanced chemical reaction is,

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From the balanced reaction we conclude that

As, 1 mole of $Ba(OH)_2$ react with 2 mole of $HCl$

So, $9.71\times 10^{-3}$ moles of $Ba(OH)_2$ react with $9.71\times 10^{-3}\times 2=0.0194$ moles of $HCl$

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From the reaction, we conclude that

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