Respuesta :
This problem is being solved using Ideal Gas Equation.
PV = nRT
Data Given:
Initial Temperature = T₁ = 0 °C = 273 K
Initial Pressure = P₁ = 101.325 KPa
Initial Volume = V₁ = 2 L
Final Temperature = T₂ = 30 °C = 303 K
Final Pressure = P₂ = 90 KPa
Final Volume = V₂ = ?
As,
Gas constant R is constant, so, Ideal gas equation can be written as,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = (P₁ × V₁ × T₂) ÷ (T₁ × P₂)
Putting Values,
V₂ = (101.325 KPa × 2 L × 303 K) ÷ (273 K × 90 KPa)
V₂ = 61402.95 ÷ 24570
V₂ = 2.499 ≈ 2.5 L
PV = nRT
Data Given:
Initial Temperature = T₁ = 0 °C = 273 K
Initial Pressure = P₁ = 101.325 KPa
Initial Volume = V₁ = 2 L
Final Temperature = T₂ = 30 °C = 303 K
Final Pressure = P₂ = 90 KPa
Final Volume = V₂ = ?
As,
Gas constant R is constant, so, Ideal gas equation can be written as,
P₁ V₁ / T₁ = P₂ V₂ / T₂
Solving for V₂,
V₂ = (P₁ × V₁ × T₂) ÷ (T₁ × P₂)
Putting Values,
V₂ = (101.325 KPa × 2 L × 303 K) ÷ (273 K × 90 KPa)
V₂ = 61402.95 ÷ 24570
V₂ = 2.499 ≈ 2.5 L
Answer : The volume of this sample is, 2.52 L
Solution : Given,
Volume of gas = 2 L
At STP,
Temperature of the gas = 273 K
Pressure of the gas = 1 atm
At temperature [tex]30^oC[/tex], the pressure is 90 kilopascal
Using ideal gas law,
[tex]PV=nRT[/tex]
(n and R is constant for this problem)
Or, [tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex]
where,
[tex]P_1[/tex] = initial pressure of gas = 1 atm
[tex]P_2[/tex] = final pressure of gas = [tex]90Kpa=\frac{90}{101.325}=0.88atm[/tex]
(1 atm = 101.325 Kpa)
[tex]V_1[/tex] = initial volume of gas = 2 L
[tex]V_2[/tex] = final volume of gas
[tex]T_1[/tex] = initial temperature of gas = 273 K
[tex]T_2[/tex] = final temperature of gas = [tex]30^oC=273+30=303K[/tex]
[tex](0^oC=273K)[/tex]
Now put all the given values in the above formula, we get the final volume of the sample.
[tex]\frac{(1atm)\times (2L)}{273K}=\frac{(0.88atm)\times V_2}{303K}[/tex]
[tex]V_2=2.52L[/tex]
Therefore, the final volume of the sample is, 2.52 L
