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Gas Laws Fact Sheet

\begin{tabular}{|l|l|}
\hline Ideal gas law & [tex]$P V=n R T$[/tex] \\
\hline \multirow{4}{*}{Ideal gas constant} & [tex]$R=8.314 \frac{L \cdot kPa}{mol \cdot K}$[/tex] \\
\hline & Or \\
\hline Standard atmospheric pressure & [tex]$R=0.0821 \frac{L \cdot atm}{mol \cdot K}$[/tex] \\
\hline Celsius to Kelvin conversion & [tex]$1 atm=101.3 kPa$[/tex] \\
\hline
\end{tabular}

Type the correct answer in the box. Express your answer to two significant figures.

The gas in an engine cylinder is compressed until it has a volume of 0.045 liters and an absolute pressure of 28 atmospheres. If the engine cylinder contains 0.020 mole of gas, what is its temperature?

The temperature of the gas is [tex]$\square$[/tex]


Sagot :

To find the temperature of the gas in an engine cylinder, we can use the Ideal Gas Law, which is mathematically represented as:

[tex]\[ PV = nRT \][/tex]

Where:
- [tex]\( P \)[/tex] is the pressure of the gas
- [tex]\( V \)[/tex] is the volume of the gas
- [tex]\( n \)[/tex] is the number of moles of gas
- [tex]\( R \)[/tex] is the Universal Gas Constant
- [tex]\( T \)[/tex] is the temperature of the gas

We are given the following values:
- Pressure ([tex]\( P \)[/tex]) = 28 atmospheres
- Volume ([tex]\( V \)[/tex]) = 0.045 liters
- Number of moles ([tex]\( n \)[/tex]) = 0.020 moles
- Gas constant in the appropriate units ([tex]\( R \)[/tex]) = 0.0821 [tex]\( \frac{\text{L atm}}{\text{mol K}} \)[/tex]

We need to find the temperature ([tex]\( T \)[/tex]). Rearranging the Ideal Gas Law equation to solve for [tex]\( T \)[/tex]:

[tex]\[ T = \frac{PV}{nR} \][/tex]

Substitute the given values into the equation:

[tex]\[ T = \frac{(28 \, \text{atm}) \times (0.045 \, \text{L})}{(0.020 \, \text{mol}) \times (0.0821 \, \frac{\text{L atm}}{\text{mol K}})} \][/tex]

Calculating the values step-by-step:

1. Calculate the numerator:

[tex]\[ P \times V = 28 \, \text{atm} \times 0.045 \, \text{L} = 1.26 \, \text{L atm} \][/tex]

2. Calculate the denominator:

[tex]\[ n \times R = 0.020 \, \text{mol} \times 0.0821 \, \frac{\text{L atm}}{\text{mol K}} = 0.001642 \, \text{L atm/K} \][/tex]

3. Divide the numerator by the denominator to find the temperature:

[tex]\[ T = \frac{1.26 \, \text{L atm}}{0.001642 \, \text{L atm/K}} = 767.356 \, \text{K} \][/tex]

Expressing the answer to two significant figures, we get:

[tex]\[ T = 767.36 \, \text{K} \][/tex]

Therefore, the temperature of the gas is [tex]\( 767.36 \, \text{K} \)[/tex].
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