A 73.8 g sample of O2 gas at 0.0 oC and 5.065x10^4 Pa is compressed and heated until the volume is 3.26 L and the temperature is 27 oC. What is the final pressure in Pa? T= 273K and 300K i think u have to convert L to m^3?

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A 73.8 g sample of O2 gas at 0.0 oC and 5.065x10^4 Pa is compressed and heated until the volume is 3.26 L and the temperature is 27 oC. What is the final pressure in Pa? T= 273K and 300K i think u have to convert L to m^3?

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Answer 1 · 2015-11-24T13:32:56

$1.8 \cdot 10^{6} Pa$ $1.8 * 10^6"Pa"$

Explanation:

Yes, you will need to do some unit conversions, but I think it'll be easier to just convert the pressure from Pa to atm.

The idea here is that the amount of gas remains unchanged, but that changing the volume and temperature of the sample will result in a change in pressure.

This means that you can use the combined gas law equation to help you find the new pressure of the sample.

$\frac{P_{1} V_{1}}{T_{1}} = \frac{P_{2} V_{2}}{T_{2}}$ $color(blue)((P_1V_1)/T_1 = (P_2V_2)/T_2)" "$ , where

$P_{1}$ $P_1$ , $V_{1}$ $V_1$ , $T_{1}$ $T_1$ - the pressure, volume, and temperature of the gas at an initial state
$P_{2}$ $P_2$ , $V_{2}$ $V_2$ , $T_{2}$ $T_2$ - the pressure, volume, and temperature of the gas at a final state

The two temperature of the gas must be expressed in Kelvin. So, rearrange the above equation to solve for $P_{2}$ $P_2$ , the new pressure of the sample

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

Before solving for $P_{2}$ $P_2$ , you need to use the ideal gas law equation to determine the initial volume of the sample, $V_{1}$ $V_1$ .

$P_{1} V_{1} = n R T_{1} \Rightarrow V_{1} = \frac{n R T_{1}}{P_{1}}$ $color(blue)(P_1V_1 = nRT_1 implies V_1 = (nRT_1)/P_1$

Use oxygen gas' molar mass to determine how many moles you have in the sample

$73.8color(red)(cancel(color(black)("g"))) * "1 mole O"_2/(32.0color(red)(cancel(color(black)("g")))) = "2.301 moles O"_2$

Now, you need to convert the initial pressure of the gas from Pa to atm by using the conversion factor

$"1 atm " = " 101325 Pa" = 1.101325 * 10^5"Pa"$

$5.065 * 10^4color(red)(cancel(color(black)("Pa"))) * "1 atm"/(1.01325 * 10^5color(red)(cancel(color(black)("Pa")))) = "0.4999 atm"$

Now plug in your values and solve for $V_$ , the initial volume of the gas

$V = (2.301color(red)(cancel(color(black)("moles"))) * 0.0821 (color(red)(cancel(color(black)("atm"))) * "L")/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 273.15color(red)(cancel(color(black)("K"))))/(0.4999color(red)(cancel(color(black)("atm")))) = "103.41 L"$

Now that you have all you need to find $P_2$ , plug in your values into the combined gas law equation - use the pressure in Pa, since you need the final pressure to be in Pa as well!

$P_2 = ( 103.41 color(red)(cancel(color(black)("L"))))/(3.26color(red)(cancel(color(black)("L")))) * ( 300.15color(red)(cancel(color(black)("K"))))/(273.15color(red)(cancel(color(black)("K")))) * 5.065 * 10^4"Pa"$

$P_2 = color(green)(1.8 * 10^6"Pa")$

The answer is rounded to two sig figs, the number of sig figs you have for the two temperatures of the gas.

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A 73.8 g sample of O2 gas at 0.0 oC and 5.065x10^4 Pa is compressed and heated until the volume is 3.26 L and the temperature is 27 oC. What is the final pressure in Pa? T= 273K and 300K i think u have to convert L to m^3?

1 Answer

Explanation:

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