Question

Two moles of a monatomic ideal gas such as helium is compressed adiabatically and reversibly from a state (3 atm, 5 L) to a state with a pressure of 4 atm.

(a) Find the volume and temperature of the final state.
(b) Find the temperature of the initial state.
(c) Find work done by the gas in the process.
(d) Find the change in internal energy in the process.
Assume CV=5R and Cp=CV+R for the diatomic ideal gas in the conditions given.

Answer 1

The final volume and temperature of the monatomic ideal gas in the adiabatic and reversible compression process are 3.75 L and 253.85 K, respectively.

Step-by-step explanation:

In an adiabatic and reversible process, the volume and temperature of a monatomic ideal gas can be determined using the adiabatic equation:

TV(γ-1) = constant, where T is the temperature, V is the volume, and γ is the ratio of specific heats (given γ = 5/3 for monatomic ideal gases).

(a) To find the final state, we can use the equation above with the given initial state (P1 = 3 atm, V1 = 5 L) and final pressure (P2 = 4 atm):

T2V2(γ-1) = T1V1(γ-1)

Solving for V2 and T2, we get:

V2 = V1(P1/P2) = 5 L (3 atm/4 atm) = 3.75 L

T2 = (T1V1(γ-1))/V2(γ-1) = (300 K * 5 L(5/3-1))/3.75 L(5/3-1) = 253.85 K