Respuesta :
Answer:
231.98 mL
Explanation:
V₁ = Initial volume = 205 mL
V₂ = Final volume
P₁ = Initial pressure = 712 mmHg
P₂ = Final pressure = 750 mmhg (STP)
T₁ = Initial temperature = -44 °C = 229.15 K
T₂ = Final temperature = 273.15 K (STP)
From Combined gas law
[tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\\\Rightarrow V_2=\frac{P_1V_1T_2}{T_1P_2}\\\Rightarrow V_2=\frac{712\times 205\times 273.15}{229.15\times 750}=231.98\ mL[/tex]
Volume of the argon at STP is 231.98 mL
Answer:
The volume of the argon at STP is 228.95 mL.
Explanation:
Given that,
Initial volume = 205 mL
Initial temperature = -44.0 °C = 229\ K
Final temperature = 273 K
Initial pressure = 712 mm Hg
Final pressure = 760 mm Hg
We need to calculate the volume of the argon at STP
Using gas law
[tex]\dfrac{P_{1}V_{1}}{T_{1}}=\dfrac{P_{2}V_{2}}{T_{2}}[/tex]
Put the value into the formula
[tex]\dfrac{712\times205}{229}=\dfrac{760\times V_{2}}{273}[/tex]
[tex]V_{2}=\dfrac{712\times205\times273}{229\times760}[/tex]
[tex]V_{2}=228.95\ mL[/tex]
Hence, The volume of the argon at STP is 228.95 mL.
