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A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure
434.32mL
Using the combined gas law:
[tex]\frac{PV}{T}[/tex] = k
Where;
P = Pressure
V = Volume
T = Temperature
k = constant.
It can be deduced that:
[tex]\frac{P_1V_1}{T_1}[/tex] = [tex]\frac{P_2V_2}{T_2}[/tex] = k ---------------------(i)
Where:
P₁ and P₂ are the initial and final pressures of the given gas
V₁ and V₂ are the initial and final volumes of the given gas
T₁ and T₂ are the initial and final temperatures of the gas.
From the question:
the gas is methane
P₁ = 760mmHg
P₂ = 616mmHg
V₁ = 398mL
V₂ = ?
T₁ = 30°C = (30 +273)K = 303K
T₂ = -5°C = (-5 +273)K = 268K
Substitute these values into equation (i) as follows;
[tex]\frac{760*398}{303}[/tex] = [tex]\frac{616*V_2}{268}[/tex]
Solve for V₂
V₂ = [tex]\frac{760*398*268}{616*303}[/tex]
V₂ = 434.32mL
Therefore, the volume of the sample at -5C and 616mmHg pressure is 434.32mL