Question

The initial state of a quantity of monatomic ideal gas is P = 1 atm, V = 1 liter, and T = 373 K. The gas is isothermally expanded to a volume of 2 liters and is then cooled at constant pressure to the volume V. This volume is such that a reversible adiabatic compression to a pressure of 1 atm returns the system to its initial state. All of the changes of state are conducted reversibly. Calculate the value of V and the total work done on or by the gas.

Answer 1

V is approximately 1.52 liters

The work done on the gas = 37 J

Step-by-step explanation:

The given information are;

Type of gas = Monoatomic gas

p₁ = 1 atm = 101325 Pa

v₁ = 1 liter = 0.001 m³

T₁ = 373 K

v₂ = 2 liters = 0.002 m³

Final volume = V

For isothermal expansion, we have, Boyle's law given as follows;

p₁×v₁ = p₂×v₂

∴ p₂ = p₁×v₁/(v₂)

p₂ = 1 atm × (1 liter)/(2 liters) = 0.5 atm = 50,662.5 Pa

We have for adiabatic compression, we have;

At V, P = p₂ = 0.5 atm (The gas is cooled at constant pressure) and can be reversed back adiabatically to p₁, v₁

Therefore we have;

`(p_1)/(p_2) = \left [(V)/(v_1) \right ]^\gamma`

γ = 1.66 for a monoatomic gas, which gives;

`(1 \ atm)/(0.5 \ atm) = \left [(V)/(1 \ liter) \right ]^(1.66)`

`V = 1 \ liter * \sqrt[1.66]{(1 \ atm)/(0.5 \ atm)} = 1 \ liter * \sqrt[1.66]{2} \approx 1.52 \ liters`

V ≈ 1.52 liters = 0.00152 m³

The total work done is given given by the following relation;

`W = (K * \left ( v_f^(1-\gamma)-v_i^(1-\gamma) \right ))/(1 - \gamma)`

`K = p * v^(\gamma ) = 0.5 * \sqrt[1.66]{2} ^(1.66 ) = 50,662.5 * (0.00152)^(1.66) \approx 1.06\ Pa \cdot m^(4.98)`

`W = \frac{1.06* \left ( 0.00152^{{1-1.66}} -0.001^(1-1.66) \right )}{1 - 1.66} \approx 36.9798 \approx 37 \ J`

Given that the work done is positive, we have that work is done in the gas

The work done on the gas = 37 J.