Respuesta :
This is an exercise in the general or combined gas law.
To start solving this exercise, we must obtain the following data:
Data:
- V₁ = 4.5 l
- T₁ = 33 °C + 273 = 306 k
- P₁ = 6.54 atm
- T₂ = 94 °C + 273 = 367 k
- V₂ = 2.3 l
- P₂ = ¿?
We use the following formula:
- P₁V₁T₂ = P₂V₂T₁ ⇒ General Formula
Where
- P₁ = Initial pressure
- V₁ = Initial volume
- T₂ = Initial temperature
- P₂ = Final pressure
- V₂ = Final volume
- T₁ = Initial temperature
We clear the general formula for the final pressure.
[tex]\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{P_{1}V_{1}T_{2} }{V_{2}T_{1}} \ \to \ Clear \ formula \end{gathered}$}[/tex]
We solve by substituting our data in the formula:
[tex]\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{(6.54 \ atm)(4.5 \not{l})(367 \not{K}) }{(2.3 \not{l})(306 \not{k})} \end{gathered}$}[/tex]
[tex]\large\displaystyle\text{$\begin{gathered}\sf P_{2}=\frac{10800.81}{ 703.8 } \ atm \end{gathered}$}[/tex]
[tex]\boxed{\large\displaystyle\text{$\begin{gathered}\sf P_{2}=15.346 \ atm \end{gathered}$} }[/tex]
If I raise the temperature to 94°C and decrease the volume to 2.3 liters, the pressure of the gas will be 15,346 atm.
Answer:
15.35 atm
Explanation:
Combined Gas Law
[tex]\sf \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}[/tex]
Temperature must be in kelvins (K).
To convert Celsius to kelvins, add 273.15.
Volume can be in any unit.
Given values:
- P₁ = 6.54 atm
- V₁ = 4.5 L
- T₁ = 33 °C = 33 + 273.15 = 306.15 K
- P₂ =
- V₂ = 2.3 L
- T₂ = 94 °C = 94 + 273.15 = 367.15 K
Rearrange the equation to isolate P₂:
[tex]\sf \implies P_2=\dfrac{P_1V_1T_2}{V_2T_1}[/tex]
Substitute the given values into the equation:
[tex]\sf \implies P_2=\dfrac{6.54 \cdot 4.5 \cdot 367.15}{2.3 \cdot 306.15}[/tex]
[tex]\sf \implies P_2=15.34516967[/tex]
Therefore, the pressure of the gas will be 15.35 atm (2 d.p.).