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3 years ago

If the temperature of a gas is raised from 30°C to 60°C, what happens to the pressure?

Chemistry

1 answer:

Fofino [41]3 years ago

5 0

<h3>Answer:</h3>

The pressure increases by 10% of the original pressure

Thus the new pressure is 1.1 times the original pressure.

<h3>Explanation:</h3>

We are given;

Initial temperature as 30°C, but K = °C + 273.15
Thus, Initial temperature, T1 =303.15 K
Final temperature, T2 is 333.15 K

We are required to state what happens to the pressure;

We are going to base our arguments to Pressure law;
According to pressure law, the pressure of a gas and its temperature are directly proportional at a constant volume

That is; P α T
Therefore, at varying pressure and temperature

$\frac{P1}{T1}=\frac{P2}{T2}$

Assuming the initial pressure, P1 is P

Rearranging the formula;

[tex]P2=\frac{P1T2}{T1}[/tex]

$P2=\frac{(P)(333.15K)}{303.15K}$

$P2 = 1.099P$

= 1.10 P

The new pressure becomes 1.10P

This means the pressure has increased by 10%

We can conclude that, the new pressure will be 1.1 times the original pressure.

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Answer : The resulting concentrations of CV and NaOH are 0.0027 M and 0.025 M respectively.

Explanation :

Step 1 : Find moles of crystal violet and NaOH.

The molarity formula is

$Molarity = \frac{mol}{L}$

Molarity of crystal violet = $3.00 \times 10^{-3} = \frac{mol (CrystalViolet)}{L}$

The volume of crystal violet solution is 18 mL which is 0.018 L.

Moles of crystal violet = $3.00 \times 10^{-3} \times 0.018 = 5.4 \times 10^{-5}$

Moles of crystal violet = 5.4 x 10⁻⁵

Moles of NaOH = $Molarity \times L = 0.250 \times 0.00200 = 5.00 \times 10^{-4}$

Moles of NaOH = 5.00 x 10⁻⁴

Step 2 : Find total volume of the solution

The total volume of the solution after mixing NaOH and crystal violet is

0.018 L + 0.00200 = 0.020 L

Step 3 : Use molarity formula to find final concentrations

Molarity of crystal violet = $\frac{mol(CrystalViolet)}{Total Volume(L) } = \frac{5.4 \times 10^{-5}}{0.020} = 2.7 \times 10^{-3}$

Final concentration of CV = 0.0027 M

Molarity of NaOH= $\frac{mol(NaOH)}{Total Volume(L) } = \frac{5.00 \times 10^{-4}}{0.020} = 0.025 \times 10^{-3}$

NaOH is a strong base and dissociates completely as follows.

$NaOH (aq) \rightarrow Na^{+} (aq) + OH^{-} (aq)$

The mole ratio of NaOH and OH⁻ is 1:1 . Therefore the concentration of OH⁻ is same as that of NaOH.

Concentration of OH⁻ = 0.025 M

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Explanation:

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109.5

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