(II) The pressure in an ideal gas is cut in half slowly, while being kept in a container with rigid walls. In the process,
465 kJ of heat left the gas. (a) How much work was done
during this process? (b) What was the change in internal
energy of the gas during this process?

(a) The work done during the given process is zero.  

(b) The required change in internal energy is -465 kJ.

Given data:

The amount of heat leaving from the gas is, Q = 465 kJ.

As per the concepts and fundamentals of thermodynamics, whenever there is some heat absorption or heat release from the system, then the volume of gas inside the system changes, this change is determined by the work done during a thermodynamic process. So, the expression for the work done is given as,

W = PV

here, P is the pressure and V is the Volume.

(a)

Since, there is no change in volume , which means V = 0.

Hence, the work done is,

W = P (0)

W = 0

Thus, we can conclude that the work done during the given process is zero.  

(b)

Applying the first law of thermodynamics as,

Q = U + W

Here, U is the Change in internal energy. Solving as,

- 465 kJ = U + 0

Negative shows that the heat is leaving from the system. So,

U = - 465 kJ.

Thus, we can conclude that the required change in internal energy is -465 kJ.

Learn more about the first law of thermodynamics here:

link

Explanation:

(a) The volume of the gas doesn't change, so no work is done.

(b) Change in internal energy = head added + work done on gas

ΔU = Q + W

ΔU = -465 kJ + 0 kJ

ΔU = -465 kJ

vapour or gas assuming atmospheriv pressure

what is the speed of a softball that travels 20m in 2secs

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