Question

An ideal monatomic gas at 300 K expands adiabatically to twice its initial volume. Assume that the process is also quasistatic. What is its final temperature?

Answer 1

Answer:

600k

Explanation:

In this problem, we need to use a gas law that relates temperature to volume. The gas law to use here is the Charles’ law.

The Charles’ law posits that temperature and volume are directly proportional, provided that the pressure is kept constant.

Mathematically:

V1/T1 = V2/T2

We are looking at getting V2, hence we can write the mathematical equation as:

T2 = V2T1/V1

Asides the fact that we know that the gas is monoatomic, we do not know its volume. Let its initial volume be v. Since it expanded adiabatically, this means that its new volume is 2v

Hence: V1 = v , V2 = 2v , T1 = 300k and T2 is ?

Substituting these values, we have the following:

T2 = (2v * 300)/v

T2 = 600k