Answer : The temperature will be 439.5 K and combined gas law is used.
Solution : Given,
Volume = 215 ml
Initial conditions,
[tex]T_1=20^oC=273+20=293K[/tex] [tex](0^oC=273K)[/tex]
[tex]P_1=1atm[/tex]
[tex]P_2=1.5atm[/tex]
Using combined gas law,
[tex]\frac{PV}{T}=constant[/tex]
where,
P = pressure of gas
V = volume of gas
T = temperature of gas
As per the problem, at constant volume the combined gas law is expressed as,
[tex]\frac{P_1}{T_1}=\frac{P_2}{T_2}[/tex]
Now put all the given values in this formula, we get
[tex]\frac{1atm}{293K}=\frac{1.5atm}{T_2}[/tex]
By rearranging the term, we get the value of [tex]T_2[/tex].
[tex]T_2=439.5K[/tex]
Therefore, the temperature will be 439.5 K and combined gas law is used.
Answer:
The Combined Gas Law
Explanation:
Let’s compare the two conditions.
p₁ = 1 atm; V₁ = 215 mL; T₁ = 20 °C
p₂ = 1.5 atm; V₂ = 215 mL; T₂= ?
The only variables are pressure and temperature.
The most appropriate law would be Gay-Lussac’s Law (p₁/T₁ = p₂/T₂), but that’s not an option.
The only other option involving p and T is the Combined Gas Law:
p₁V₁/T₁ = p₂V₂/T₂
Since V₂ = V₁, we can write
p₁V₁/T₁ = p₂V₁/T₂ Divide both sides by V₁
p₁/T₁= p₂/T₂ Multiply both sides by T₂
p₁T₂/T₁ = p₂ Multiply both sides by T₁
p₁T₂= p₂T₁ Divide both sides by p₁
T₂ = T₁ × p₂/p₁
Now, convert your temperature to kelvins, insert the values into the formula, and calculate the new temperature.
