Question

A gas cylinder holds 0.10 mol of O2 at 150 C and a pressure of 3.0 atm. The gas expands adiabatically until the pressure is halved
Part A
What is the final volume?
Part B
What is the final temperature?

Answer 1

Answer:

V2 = 1.899*10^-3 m^3

T2 = 347.125 K

Explanation:

Using gas law, we know that

PV = nRT,

Where

V1 = 0.00115743 m^3.

gamma = 1.4

Now, when we solve for final volume, V2 we get

V2 = V1/((P2/P1)^(1/gamma))

V2 = 1.899*10^-3 m^3

Using the same law and method, when we try to solve for the temperature, we find that the final temperature, T2 is

T2 = T1*((V1/V2)^(gamma-1))

T2 = 347.125 K

Answer 2

The final volume is 1.899*10^-3 m^3

And, the final temperature is 347.125 K

Gas law:

here we used gas law,

we know that

PV = nRT,

Here

V1 = 0.00115743 m^3.

gamma = 1.4

Now final volume is

V2 = V1/((P2/P1)^(1/gamma))

V2 = 1.899*10^-3 m^3

Now the final temperature is

T2 = T1*((V1/V2)^(gamma-1))

T2 = 347.125 K

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