chemistry volume of gas

Question

I have an unknown volume of gas at a pressure of 0.50 atm and a temperature of 325 K. If I raise the pressure to 1.2 atm, decrease the temperature to 320 K, and measure the final volume to be 48 liters, what was the initial volume of the gas?

Answer 1

We can figure out the initial volume backwards from the final conditions using the ideal gas law (P*V=n*R*T).

We want to use the ideal gas constant that corresponds to the given units. R=0.08206 L atm /(mol K).

First we should solve for n, the number of moles using the formula n= P*V / (R*T)

Using the final conditions of P_F = 1.2 atm, V_F = 48 Liters, and T_F = 320 K, we get n = 2.19 moles.

It's important to know that for a closed, non-reacting volume of a gas, the number of moles, n, is a constant no matter how the conditions change. So, now that we know n, we can solve for the initial volume by rearranging the ideal gas law again. V= (n*R*T)/P.

Using the initial conditions of T_I = 325 K, P_I = 0.5 atm, and n= 2.19 moles, we get V_I = 117 L.

Another easier and quicker method of solving this uses the fact that both the ideal gas constant and n are constants. From the ideal gas law this means P*V / T = n*R = constant. Therefore P_I*V_I / T_I = P_F*V_F / T_F.

We rearrange to solve for initial volume: V_I = P_F*V_F*T_I / (P_I*T_F)

Using this method we get the same answer V_I=117 L, without having to solve for n, the number of moles or using the ideal gas constant. Hope this helps!