Question

An ideal monatomic gas expands isothermally from 0.540 m3 to 1.25 m3 at a constant temperature of 720 K. If the initial pressure is 1.20e5 Pa.

a) Find the work done on the gas
b) Find the thermal energy transfer Q
c) Find the change in the internal energy

Answer 1

a) The work done on the gas during an isothermal process is given by:

W = nRT ln(Vf/Vi)

where:
- n is the number of moles of gas
- R is the gas constant (8.31 J/mol*K)
- T is the temperature of the gas
- Vi and Vf are the initial and final volumes of the gas, respectively

Since the gas is monatomic, its molar specific heat at constant volume is Cv = (3/2)R, and its molar specific heat at constant pressure is Cp = (5/2)R. Since the process is isothermal, the temperature of the gas remains constant, so T = 720 K for both Vi and Vf. Therefore, we can simplify the equation for work to:

W = nRT ln(Vf/Vi) = nRT ln(1.25/0.540)

We can calculate the number of moles of gas using the ideal gas law:

PV = nRT

n = PV/RT

Substituting the given values, we get:

n = (1.20 x 10^5 Pa)(0.540 m^3)/(8.31 J/mol*K)(720 K) ≈ 9.86 mol

Therefore, the work done on the gas is:

W = nRT ln(1.25/0.540) ≈ 9.92 x 10^3 J

b) The thermal energy transfer Q during an isothermal process is equal to the work done on the gas:

Q = W ≈ 9.92 x 10^3 J

c) The change in internal energy ΔU of an ideal gas during an isothermal process is zero, since the temperature of the gas remains constant and internal energy is a function of temperature only. Therefore, ΔU = 0.

Answer 2

a) The work done on the gas during an isothermal expansion is given by:

W = nRT ln(V2/V1)

where n is the number of moles of gas, R is the gas constant, T is the temperature, V1 is the initial volume, and V2 is the final volume.

Since the gas is monatomic, we can use the ideal gas law to find the number of moles:

PV = nRT

n = PV/RT

Substituting this expression for n into the equation for work, we get:

W = PV ln(V2/V1)

where we have cancelled out the R and T terms.

Substituting the given values, we get:

W = (1.20e5 Pa)(0.540 m^3) ln(1.25/0.540) = 1.38e4 J

b) The thermal energy transfer Q during an isothermal process is equal to the work done on the gas. Therefore, Q = 1.38e4 J.

c) The change in internal energy ΔU of a gas during an isothermal process is zero, since the temperature of the gas does not change. Therefore, ΔU = 0.