If 0.52L of gas at 345K is expanded to 0.763L, what will the new temperature be?

Question

I need the answer ASAP!!

Answer 1

The Ideal Gas Law says PV = nRT where P is the pressure, V is the volume, n is the number of moles, R is the Ideal Gas Constant, and T is the temperature. We can take that equation and solve it for "nR" by dividing both sides by "T" to get:

nR = PV/T

If you enclose a gas into a sealed container then the number of moles (n) is a constant (as is "R" the Ideal Gas Constant). That means, comparing Pressure, Volume and Temperature time "1" vs time 2, we can say:

P₁V₁/T₁ = nR = P₂V₂/T₂ or

P₁V₁/T₁ = P₂V₂/T₂

Since you do not mention pressure in your problem, I'm going to assume the pressure in this problem is assumed to be constant. If pressure is the same at time 1 and time 2 then the equation simplifies to:

V₁/T₁ = V₂/T₂

We are looking for T₂ so we can algebraically manipulate the equation to solve for T₂ getting:

T₂ = T₁•V₂/V₁

Before we plug in the numbers we need to ensure temperature is in Kelvin, which it is so:

T₂ = T₁•V₂/V₁

T₂ = (345)•(0.763)/(0.52) = 506 Kelvin