alka8q7

2021-11-17

The air in a bicycle tire is bubbled through water and collected at ${25}^{\circ }C$. If the total volume of gas collected is 5.45 L at a temperature of ${25}^{\circ }C$ and a pressure of 745 torr, how many moles of gas were in the bicycle tire?

Froldigh

Step 1
$T=\left(25+273.15\right)K=298.15K$
${P}_{\to t}=745\to rr=745mmHg$
${P}_{{H}_{2}O}=23.78mmHg$
$V=5.45L$
${n}_{gas}=?$
Write down the known and unknown values.
Step 2
${P}_{\to t}={P}_{{H}_{2}O}+{P}_{gas}$
${P}_{gas}={P}_{\to t}-{P}_{{H}_{2}O}$
$=745mmHg--23.7mmHg$
$=721.3mmHg$
To determine the evolved gas's partial pressure, use the gas-evolution equation based on Dalton's Law of
Step 3

Convert the evolved gas's partial pressure into atmospheres.

Step 4

To determine how many moles of the evolved gas there are, use the Ideal Gas Law.

Jazz Frenia

To solve this problem, we can use the ideal gas law equation:
$PV=nRT$
Where:
- $P$ is the pressure of the gas (in atm)
- $V$ is the volume of the gas (in liters)
- $n$ is the number of moles of gas
- $R$ is the ideal gas constant (0.0821 L·atm/mol·K)
- $T$ is the temperature of the gas (in Kelvin)
First, let's convert the given pressure from torr to atm. We know that 1 atm is equal to 760 torr. Therefore:
$P=\frac{745\phantom{\rule{0.167em}{0ex}}\text{torr}}{760\phantom{\rule{0.167em}{0ex}}\text{torr/atm}}=0.979\phantom{\rule{0.167em}{0ex}}\text{atm}$
Next, let's convert the given volume to liters:
$V=5.45\phantom{\rule{0.167em}{0ex}}\text{L}$
Now, we can rearrange the ideal gas law equation to solve for the number of moles:
$n=\frac{PV}{RT}$
Substituting the given values:
$n=\frac{\left(0.979\phantom{\rule{0.167em}{0ex}}\text{atm}\right)\left(5.45\phantom{\rule{0.167em}{0ex}}\text{L}\right)}{\left(0.0821\phantom{\rule{0.167em}{0ex}}\text{L·atm/mol·K}\right)\left(25+273.15\right)\phantom{\rule{0.167em}{0ex}}\text{K}}$
$n=\frac{5.34\phantom{\rule{0.167em}{0ex}}\text{L·atm}}{22.42\phantom{\rule{0.167em}{0ex}}\text{L·atm/mol·K}}$
$n=0.237\phantom{\rule{0.167em}{0ex}}\text{mol}$
However, we need to account for the fact that the gas was collected at 25 degrees Celsius, not Kelvin. To convert the temperature from Celsius to Kelvin, we add 273.15:
$n=0.237\phantom{\rule{0.167em}{0ex}}\text{mol}×\frac{298.15\phantom{\rule{0.167em}{0ex}}\text{K}}{273.15\phantom{\rule{0.167em}{0ex}}\text{K}}$
$n=0.260\phantom{\rule{0.167em}{0ex}}\text{mol}$
Therefore, the number of moles of gas in the bicycle tire is approximately 0.260 mol, rounded to three decimal places.

Andre BalkonE

Given:
Total volume of gas collected, $V=5.45\phantom{\rule{0.167em}{0ex}}\text{L}$
Temperature, $T=25\phantom{\rule{0.167em}{0ex}}\text{°C}$
Pressure, $P=745\phantom{\rule{0.167em}{0ex}}\text{torr}$
To find the number of moles of gas, we can use the ideal gas law equation:
$PV=nRT$
First, we need to convert the temperature from Celsius to Kelvin. We use the equation:
$T\left(K\right)=T\left(°C\right)+273.15$
Plugging in the given values:
$T=25\phantom{\rule{0.167em}{0ex}}\text{°C}+273.15=298.15\phantom{\rule{0.167em}{0ex}}\text{K}$
Now, we can rearrange the ideal gas law equation to solve for the number of moles:
$n=\frac{PV}{RT}$
Substituting the given values:

Simplifying the equation gives us:
$n=0.211\phantom{\rule{0.167em}{0ex}}\text{mol}$
Therefore, there were 0.211 moles of gas in the bicycle tire.
Hence, the solution to the problem is $0.211$ mol.

Mr Solver

First, we need to convert the temperature from Celsius to Kelvin using the equation:
$T\left(K\right)=T\left(°C\right)+273.15$
Plugging in the values, we get:
$T\left(K\right)=25+273.15=298.15$
Now we can rearrange the ideal gas law equation to solve for $n$:
$n=\frac{PV}{RT}$
Plugging in the values, we get:
$n=\frac{\left(745\phantom{\rule{0.167em}{0ex}}\text{torr}\right)×\left(5.45\phantom{\rule{0.167em}{0ex}}\text{L}\right)}{\left(0.0821\phantom{\rule{0.167em}{0ex}}\text{L·atm/(mol·K)}\right)×\left(298.15\phantom{\rule{0.167em}{0ex}}\text{K}\right)}$
Simplifying the equation, we find:
$n=\frac{4061.75}{24.457\phantom{\rule{0.167em}{0ex}}\text{mol}}\approx 166.26\phantom{\rule{0.167em}{0ex}}\text{mol}$
Therefore, there were approximately 166.26 moles of gas in the bicycle tire.
(Note: We should round the final answer to the appropriate number of significant figures based on the given data. In this case, since the total volume of gas collected is given with three significant figures, the final answer should be rounded to three significant figures as well.)

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