Respuesta :
Answer:
[tex]T=29.2326 \ K[/tex]
Explanation:
We can use the ideal gas law to answer this question. The ideal gas law relates a gasses pressure, volume, and temperature and is written as follows.
[tex]\boxed{\left\begin{array}{ccc}\text{\underline{The Ideal Gas Law:}}\\\\PV=nRT\end{array}\right}[/tex]
"n" is the number of moles present in the gas and "R" is referred to as the universal gas constant.
[tex]R=0.0821 \ \frac{atm \cdot L}{mol \cdot K} \ \text{or} \ 8.31 \ \frac{J}{mol \cdot K}[/tex]
Be careful when using the ideal gas law, make sure to use the appropriate R value and remember that T is measured in kelvin.
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Given:
[tex]P=1.30 \ atm\\V=2.40 \ L\\n=1.30 \ mol\\R=0.0821 \ \frac{atm \cdot L}{mol \cdot K} \[/tex]
Find:
[tex]T= \ ?? \ K[/tex]
(1) - Solve the ideal gas law for "T"
[tex]PV=nRT\\\\\Longrightarrow T=\frac{PV}{nR}[/tex]
(2) - Plug the known values into the equation
[tex]T=\frac{PV}{nR} \\\\\Longrightarrow T=\frac{(1.30)(2.40)}{(1.30)(0.0821)} \\\\\therefore \boxed{\boxed{T=29.2326 \ K}}[/tex]
Thus, the gasses temperature is found.
To determine the temperature at which 1.30 mole of an ideal gas in a 2.40 L container exerts a pressure of 1.30 atm, we can use the ideal gas law equation: PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
We can rearrange the equation to solve for temperature (T):
T = PV / (nR)
Given:
P = 1.30 atm
V = 2.40 L
n = 1.30 mole
R = ideal gas constant (8.314 J/(mol·K))
Substituting the values into the equation:
T = (1.30 atm) * (2.40 L) / (1.30 mole * 8.314 J/(mol·K))
T ≈ 2.56 K
Therefore, at approximately 2.56 Kelvin, 1.30 mole of the ideal gas in a 2.40 L container will exert a pressure of 1.30 atm.
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