A gas occupies ${67. cm}^{3}$ $"67. cm"^3$ at $9.38 \times 10^{4} l Pa$ $9.38 × 10^4color(white)(l)"Pa"$ and 22 °C. What is its volume at $10.6 \times 10^{5} l Pa$ $10.6 × 10^5color(white)(l)"Pa"$ and 29 °C?

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A gas occupies ${67. cm}^{3}$ $"67. cm"^3$ at $9.38 \times 10^{4} l Pa$ $9.38 × 10^4color(white)(l)"Pa"$ and 22 °C. What is its volume at $10.6 \times 10^{5} l Pa$ $10.6 × 10^5color(white)(l)"Pa"$ and 29 °C?

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Answer 1 · 2015-11-03T00:35:05

${6.1 cm}^{3}$ $"6.1 cm"^3$

Explanation:

So, it's always a good idea to start by making a note of what information is being provided by the problem.

In your case, you know that the initial sample of gas

occupies a volume equal to ${67 cm}^{3}$ $"67 cm"^3$

has a temperature of $22^{\circ} C$ $22^@"C"$

has a pressure of $9.38 \cdot 10^{4} Pa$ $9.38 * 10^4"Pa"$

You then go on to change the temperature to $29^{\circ} C$ $29^@"C"$ and the pressure to $10.6 \cdot 10^{5} Pa$ $10.6 * 10^5"Pa"$ .

Notice that no mention of number of moles was made. This means that you can assume it to be constant. So, if you start from the ideal gas law equation, you can say that

$P_{1} \cdot V_{1} = n \cdot R \cdot T_{1} \to$ $P_1 * V_1 = n * R * T_1 ->$ the initial state of the gas

and

$P_{2} \cdot V_{2} = n \cdot R \cdot T_{2} \to$ $P_2 * V_2 = n * R * T_2 ->$ the final state of the gas

Since $n$ $n$ is constant, and $R$ $R$ is the universal gas constant, you can rearrange these equations to isolate these two constant terms on one side

$\frac{P_{1} \cdot V_{1}}{T_{1}} = n \cdot R$ $(P_1 * V_1)/T_1 = n * R" "$ and $\frac{P_{2} \cdot V_{2}}{T_{2}} = n \cdot R$ $" "(P_2 * V_2)/T_2 = n * R$

Notice that you have two expressions that are equal to the same value, $n \cdot R$ $n * R$ . This means that they are equal to each other as well.

$\frac{P_{1} \cdot V_{1}}{T_{1}} = \frac{P_{2} \cdot V_{2}}{T_{2}} \to$ $(P_1 * V_1)/T_1 = (P_2 * V_2)/T_2 ->$ the combined gas law equation

Now all you have to do is rearrange this to solve for $V_{2}$ $V_2$ , the volume of the gas at the final state.

Look what happens if you divide both sides of the equation by $P_{2}$ $P_2$

$\frac{P_{1}}{P_{2}} \cdot \frac{V_{1}}{T_{1}} = \frac{P_{2} \cdot V_{2}}{T_{2} \cdot P_{2}}$ $P_1/P_2 * V_1/T_1 = (color(red)(cancel(color(black)(P_2))) * V_2)/(T_2 * color(red)(cancel(color(black)(P_2))))$

$\frac{P_{1}}{P_{2}} \cdot \frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $P_1/P_2 * V_1/T_1 = V_2/T_2$

Now multiply both sides by $T_{2}$ $T_2$ to get $V_{2}$ $V_2$ alone one one side of the equation

$\frac{P_{1}}{P_{2}} \cdot \frac{T_{2}}{T_{1}} \cdot V_{1} = \frac{V_{2}}{T_{2}} \cdot T_{2}$ $P_1/P_2 * T_2/T_1 * V_1 = V_2/color(red)(cancel(color(black)(T_2))) * color(red)(cancel(color(black)(T_2)))$

Finally, you got

$V_{2} = \frac{P_{1}}{P_{2}} \cdot \frac{T_{2}}{T_{1}} \cdot V_{1}$ $V_2 = P_1/P_2 * T_2/T_1 * V_1$

Now plug in your values and solve for $V_{2}$ $V_2$ - do not foget to convert the temperature from degrees Celsius to Kelvin!

$V_{2} = \frac{9.38 \cdot 10^{4} Pa}{10.6 \cdot 10^{5} Pa} \cdot \frac{(273.15 + 29) K}{(273.15 + 22) K} \cdot {67 cm}^{3}$ $V_2 = (9.38 * 10^4color(red)(cancel(color(black)("Pa"))))/(10.6 * 10^5color(red)(cancel(color(black)("Pa")))) * ((273.15 + 29)color(red)(cancel(color(black)("K"))))/((273.15 + 22)color(red)(cancel(color(black)("K")))) * "67 cm"^3$

$V_{2} = {6.0695 cm}^{3}$ $V_2 = "6.0695 cm"^3$

You need to round this off to two sig figs, the number of sig figs you have for the initial volume of the gas

$V_{2} = {6.1 cm}^{3}$ $V_2 = color(green)("6.1 cm"^3)$

Answer 2 · 2015-11-03T01:53:06

The new volume will be ${6.1 cm}^{3}$ $color(blue)("6.1 cm"^3)$ .

Explanation:

We use the Combined Gas Law equation,

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $(P_1V_1)/T_1 = (P_2V_2)/T_2$

Let's start by listing our given information.

$P_{1} = 9.38 \times 10^{4} Pa$ $P_1 = 9.38 × 10^4 "Pa"$ ; $V_{1} = {67. cm}^{3}$ $V_1 = "67. cm"^3$ ; $T_{1} = 22 °C =(22 + 273.15) K = 295 K$ $T_1 = "22 °C =(22 + 273.15) K = 295 K"$
$P_{2} = 10.6 \times 10^{5} Pa$ $P_2 = 10.6 × 10^5 "Pa"$ ; $V_{2} = ?$ $V_2 = "?"$ ; $T_{2} = 29 °C = (29 + 273.15) K = 302 K$ $T_2 = "29 °C = (29 + 273.15) K = 302 K"$

Now we must rearrange the Combined Gas Law equation to get $V_{2}$ $V_2$ by itself.

We'll take it step by step.

Step 1. Multiply both sides by $T_{2}$ $T_2$ .

$\frac{P_{1} V_{1}}{T_{1}} \times T_{2} = \frac{P_{2} V_{2}}{T_{1}} \times T_{2}$ $(P_1V_1)/T_1 × T_2 = (P_2V_2)/(color(red)(cancel(color(black)(T_1)))) × color(red)(cancel(color(black)(T_2)$

$\frac{P_{1} V_{1} T_{2}}{T_{1}} = P_{2} V_{2}$ $(P_1V_1T_2)/T_1 = P_2V_2$

Step 2. Divide both sides by $P_{2}$ $P_2$ .

$\frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} = \frac{P_{2} V_{2}}{P_{2}}$ $(P_1V_1T_2)/(P_2T_1) = (color(red)(cancel(color(black)(P_2)))V_2)/color(red)(cancel(color(black)(P_2)$

$\frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} = V_{2}$ $(P_1V_1T_2)/(P_2T_1) = V_2$ or $V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}}$ $V_2 = (P_1V_1T_2)/(P_2T_1)$

Now we insert the values into the equation.

$V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}} = \frac{9.38 \times 10^{4} Pa \times {67. cm}^{3} \times 302 K}{10.6 \times 10^{5} Pa \times 295 K} = {6.1 cm}^{3}$ $V_2 = (P_1V_1T_2)/(P_2T_1) = (9.38 × 10^4 color(red)(cancel(color(black)("Pa"))) × "67. cm"^ 3 × 302 color(red)(cancel(color(black)("K"))))/( 10.6 × 10^5 color(red)(cancel(color(black)("Pa"))) × 295 color(red)(cancel(color(black)("K")))) = "6.1 cm"^3$

Check: The temperature doesn't change much, but the pressure increases by about ten-fold.

The new volume should be about one-tenth of the original volume, or about ${7 cm}^{3}$ $"7 cm"^3$ .

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A gas occupies ${67. cm}^{3}$ $"67. cm"^3$ at $9.38 \times 10^{4} l Pa$ $9.38 × 10^4color(white)(l)"Pa"$ and 22 °C. What is its volume at $10.6 \times 10^{5} l Pa$ $10.6 × 10^5color(white)(l)"Pa"$ and 29 °C?

2 Answers

Explanation:

Explanation:

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A gas occupies ${67. cm}^{3}$ $"67. cm"^3$ at $9.38 \times 10^{4} l Pa$ $9.38 × 10^4color(white)(l)"Pa"$ and 22 °C. What is its volume at $10.6 \times 10^{5} l Pa$ $10.6 × 10^5color(white)(l)"Pa"$ and 29 °C?