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.
[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.