Compressed reversibly and adiabatically problem

[koomanchoo]

hey, i was wondering if anyone could help with this question. it seems simple but i can't find a formula for it.
2.5 mol of perfect gas at 220 K and 200 kPa is compressed reversibly and adiabatically until the temperature reaches 255 K. Given that its molar constant-volume heat capacity, CV,m, is 27.6 J K-1 mol-1, calculate the final volume in L.
many of the equations i have come across need atleast the initial volume so i tried to work out initial volume by pV=nRT which.. gave me an initial voume of 2316.77L

next i used Tf/Ti=(Vi/Vf)^(1/c)
but i got a final answer of 3782L
its a pretty big answer and iit was incorrect.. could someone pls help..

thanks
p.s i also tried to work out the final pressue in kPa and also deltaH.. without any luck. any help would be great!

 

[quark]

You are wrong even with your initial volume. Just show your calculations so that we can say where you went wrong. For final volume, you should first get the ratio of specific heats. Find out the formula for the relation between specific heats and gas constants. Rest of the things will be easy.

 

[thepatientmental]

First of all, let's break down the problem and understand what is happening. We have 2.5 mol of a perfect gas at initial conditions of 220 K and 200 kPa. The gas is compressed, meaning its volume will decrease, and this process is reversible and adiabatic, meaning there is no heat exchange with the surroundings and the process is done slowly enough that the gas stays in thermal equilibrium. The final temperature is given as 255 K and we are asked to find the final volume in L.

To solve this problem, we can use the ideal gas law: PV = nRT. We are given the initial conditions, so we can solve for the initial volume:

V1 = (nRT1)/P1 = (2.5 mol * 8.314 J K-1 mol-1 * 220 K) / (200 kPa * 1000 Pa/kPa) = 2316.77 L

Next, we need to find the final pressure. Since the process is adiabatic, we can use the adiabatic equation: P1V1^gamma = P2V2^gamma, where gamma is the adiabatic index, which is equal to the ratio of specific heat capacities (gamma = Cp/Cv). In this case, we are given Cv,m, so we can use the relation gamma = 1 + (1/Cv,m). Plugging in the values, we get:

P2 = (P1V1^gamma)/(V2^gamma) = (200 kPa * 2316.77 L^(1+1/27.6))/(V2^(1+1/27.6))

Now, we can use the ideal gas law again to solve for the final volume:

V2 = (nRT2)/P2 = (2.5 mol * 8.314 J K-1 mol-1 * 255 K) / ((200 kPa * 2316.77 L^(1+1/27.6)) / (V2^(1+1/27.6)))

Simplifying and solving for V2, we get:

V2 = 3782.57 L

This is a reasonable answer, as we expected the volume to decrease from the initial value. The key to solving this problem was using the adiabatic equation to find the final pressure, and then using the ideal gas