Work under constant temperature

[ymehuuh]

Homework Statement

An ideal monatomic gas expands isothermally from 0.590 m3 to 1.25 m3 at a constant temperature of 780 K. If the initial pressure is 1.20x10^5 Pa.
(a) Find the work done on the gas.
(b) Find the thermal energy transfer Q.
(c) Find the change in the internal energy.

Homework Equations

W= -PV
deltaU=Q-W
PfVf/PiVi = Tf/Ti

The Attempt at a Solution

I know I need to find the final pressure...but I'm not sure how to.
Do I use Pf*1.35/1.2x10^5 = 780k?

 

[Andrew Mason]

Homework Statement

An ideal monatomic gas expands isothermally from 0.590 m3 to 1.25 m3 at a constant temperature of 780 K. If the initial pressure is 1.20x10^5 Pa.
(a) Find the work done on the gas.

If T is constant and PV=nRT, what is the relationship between P₁V₁ and P₂V₂? Does that help you to find P₂?

AM

 

[nlightner]

I would approach this problem by first acknowledging that the work is being done under constant temperature, meaning that the temperature of the gas remains constant throughout the expansion process. This implies that the ideal gas law (PV = nRT) can be simplified to P1V1 = P2V2, where P1 and V1 represent the initial pressure and volume, and P2 and V2 represent the final pressure and volume.

(a) To find the work done on the gas, we can use the equation W = -PΔV, where ΔV is the change in volume. In this case, ΔV = V2 - V1 = 1.25 m^3 - 0.590 m^3 = 0.660 m^3. Therefore, the work done on the gas is W = -(1.20x10^5 Pa)(0.660 m^3) = -79,200 J.

(b) To find the thermal energy transfer Q, we can use the first law of thermodynamics, which states that ΔU = Q - W, where ΔU is the change in internal energy. Since the temperature remains constant, ΔU = 0 and thus Q = W. Therefore, Q = 79,200 J.

(c) To find the change in internal energy, we can use the ideal gas law to calculate the final pressure, P2 = P1V1/V2 = (1.20x10^5 Pa)(0.590 m^3)/(1.25 m^3) = 56,640 Pa. We can then use the equation ΔU = nCvΔT, where n is the number of moles of gas, Cv is the molar specific heat at constant volume, and ΔT is the change in temperature. Since the temperature remains constant, ΔT = 0 and thus ΔU = 0. This makes sense because the internal energy of an ideal gas only depends on its temperature, which remains constant in this case.