Question

An ideal gas, at a pressure of 1.00 x 10^5 Pa, a temperature of 20.0 C, and volume of 2.00 m^3, has its volume cut in half. If the gas gains 1.00 x 10^5 J of internal energy, what type of process occurred?(A) Adiabatic(B) Isobaric (C) Isovolumetric(D) Isothermal

Answer 1

(A) Adiabatic

Step-by-step explanation:

By the first law of thermodynamics

ΔU = Q + W

Where ΔU is the change in the internal energy, Q is the heat and W is the work done to the system.

Additionally, the work done to the system can be calculated as:

W = P x ΔV

Where P is the pressure and ΔV is the change in volume.

If the ideal gas has its volume cut in half,

ΔV = 2.00 m³ - 1.00 m³ = 1.00 m³

Then, the work done to the system will be:

W = 1.00 x 10⁵ Pa x 1.00 m³

W = 1.00 x 10⁵ J.

Now, we can see that the work done to the system is equal to the change in the internal energy, this is:

W = 1.00 x 10⁵ J and U = 1.00 x 10⁵ J

By the first law of thermodynamics, Q = 0, which means that the process that occurred is Adiabatic.