Gas Laws Fact Sheet

\begin{tabular}{|l|l|}
\hline Ideal gas law & [tex]$P V=n R T$[/tex] \\
\hline \multirow{3}{*}{ Ideal gas constant } & [tex]$R=8.314 \frac{L kPa}{mol K}$[/tex] \\
& or \\
& [tex]$R=0.0821 \frac{L atm}{mol K}$[/tex] \\
\hline Standard atmospheric pressure & [tex]$1 atm=101.3 kPa$[/tex] \\
\hline Conversion to Kelvin & [tex]$T_{K}=T_{C} + 273.15$[/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 liter 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] K.



Answer :

To determine the temperature of the gas in the engine cylinder, we can use the Ideal Gas Law equation, which is given by:

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

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

Given the values:
- Pressure ([tex]\( P \)[/tex]) = 28 atmospheres,
- Volume ([tex]\( V \)[/tex]) = 0.045 liters,
- Number of moles ([tex]\( n \)[/tex]) = 0.020 moles,
- Ideal Gas Constant ([tex]\( R \)[/tex]) = 0.0821 [tex]\(\frac{L \, atm}{mol \, K}\)[/tex],

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

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

Substituting the known values into the equation:

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

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

[tex]\[ T \approx 767.3568818514007 \, K \][/tex]

When expressing the temperature to two significant figures:

[tex]\[ T \approx 770 \, K \][/tex]

Therefore, the temperature of the gas is [tex]\( \boxed{770} \)[/tex] K.