pls.. help me with this i really need it.. pls.

A gas in a cylinder expands from a volume of 0.110 m3 to 0.320 m3. Heat flows into the gas just rapidly enough to keep the pressure constant at 1.65 x 105 Pa during the expansion. The total heat added is 1.15 x 105 J. (a) Find the work done by the gas. (b) Find the change in internal energy of the gas. (c) Does it matter whether the gas is ideal? Why or why not?

tryru

Sure, I can help you with that. Let's go step by step.

(a) To find the work done by the gas, we can use the equation:

Work = Pressure x Change in Volume

For this problem, the pressure is constant at 1.65 x 10^5 Pa, and the change in volume is given as 0.320 m^3 - 0.110 m^3 = 0.210 m^3.

Substituting these values into the equation, we get:

Work = (1.65 x 10^5 Pa) x (0.210 m^3)
Work = 34,650 J

Therefore, the work done by the gas is 34,650 J.

(b) The change in internal energy can be found using the first law of thermodynamics:

Change in Internal Energy = Heat added - Work

In this problem, the heat added is given as 1.15 x 10^5 J, and we just calculated the work done as 34,650 J.

Substituting these values into the equation, we get:

Change in Internal Energy = (1.15 x 10^5 J) - (34,650 J)
Change in Internal Energy = 80,350 J

Therefore, the change in internal energy of the gas is 80,350 J.

(c) Whether the gas is ideal or not does not affect the calculations for (a) and (b) in this problem. These calculations are based on the assumptions of constant pressure and the first law of thermodynamics, which are valid for both ideal and non-ideal gases.

However, if the problem involved calculations using the ideal gas law or other properties specific to ideal gases, then it would matter whether the gas is ideal or not.

I hope this helps! Let me know if you have any further questions.

To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system:

ΔU = Q - W

(a) To find the work done by the gas, we can use the equation for work done in a gas expansion:

W = PΔV

where P is the pressure and ΔV is the change in volume. Since the pressure is constant at 1.65 x 10^5 Pa and the gas expands from a volume of 0.110 m^3 to 0.320 m^3, we can substitute these values into the equation:

W = (1.65 x 10^5 Pa)(0.320 m^3 - 0.110 m^3)

Calculating the value of W gives us:

W ≈ 3.3 x 10^4 J

Therefore, the work done by the gas is approximately 3.3 x 10^4 J.

(b) To find the change in internal energy of the gas, we can use the first law of thermodynamics equation:

ΔU = Q - W

Given that the total heat added to the gas is 1.15 x 10^5 J, we can substitute this value along with the work done into the equation:

ΔU = (1.15 x 10^5 J) - (3.3 x 10^4 J)

Calculating the value of ΔU gives us:

ΔU ≈ 8.2 x 10^4 J

Therefore, the change in internal energy of the gas is approximately 8.2 x 10^4 J.

(c) Whether the gas is ideal or not does matter. In an ideal gas, the internal energy depends only on the temperature and is independent of the volume or pressure. However, for a non-ideal gas, the internal energy can also depend on the specific properties of the gas. In this problem, we are not given any specific information about the gas or its properties, so we cannot determine whether it is ideal or non-ideal. However, the calculations for work and change in internal energy are the same regardless of whether the gas is ideal or not.

(A):

W = √( p ∆V)
For a constant pressure process, you can factor it out and get
W = p ∙ √ ∆V
= p ∙ (V_final - V_initial)
so
w = 1.65 x 10^5 (.320-.110)
W= 34650
(B): U = Q + W
(C): no