The volume of a gas is 11.8 mL at 49°C and .942 atm. What will the volume be at 17°C and 0.993 atm? Answer in units of mL

Question

Stuck on this for a long time

Answer 1

Hello Justice,

I assume we are dealing with an ideal gas. One common form of the Ideal Gas Law is

(Eq.1) PV = nRT

where the value of R (the ideal gas constant) depends on the units being used. Before proceeding, it is important to recall that T must be in Kelvin. The statement of your problem implies that n (the amount of gas measured in moles) is constant. Note that the Ideal Gas Law can be rewritten as

PV/T = nR

If n is constant, we have

PV/T = C

where C is a constant (whose value in any given problem will depend on n.) This expresses that fact that, if n is constant, then P is directly proportional to T and inversely proportional to V. This fact can also be expressed as

(Eq.2) P₁V₁/T₁ = P₂V₂/T₂

where P₁,V₁, and T₁ are the initial pressure, volume, and temperature, respectively, and P₂,V₂, and T₂ are the corresponding final values. Before completing the problem, let us convert the temperatures to Kelvin. The relationship between temperatures (in Kelvin) and temperatures in degrees Celsius is

T_K = T_C + 273.15

If T₁ = 49°C and T₂ = 17°C, then, on the Kelvin scale, we have

T₁ = 49 + 273.15 = 322.15K, and

T₂ = 17 + 273.15 =290.15K

Finally, we can use Eq. 2 to find the final volume V₂.

(.942atm)(11.8mL)/322.15K = (.993atm)V₂/(290.15K)

which gives

V₂ ≅ 10.1mL

Hope that helps! Let me know if you have any further questions.

William