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

Answer:

V is approximately 1.52 liters

The work done on the gas = 37 J

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;

$\dfrac{p_1}{p_2} = \left [\dfrac{V}{v_1} \right ]^\gamma$

γ = 1.66 for a monoatomic gas, which gives;

$\dfrac{1 \ atm}{0.5 \ atm} = \left [\dfrac{V}{1 \ liter} \right ]^{1.66}$

$V = 1 \ liter \times \sqrt[1.66]{\dfrac{1 \ atm}{0.5 \ atm}} = 1 \ liter \times \sqrt[1.66]{2} \approx 1.52 \ liters$

V ≈ 1.52 liters = 0.00152 m³

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

$W = \dfrac{K \times \left ( v_f^{1-\gamma}-v_i^{1-\gamma} \right )}{1 - \gamma}$

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

$W = \dfrac{1.06\times \left ( 0.00152^{{1-1.66}} -0.001^{1-1.66} \right )}{1 - 1.66} \approx 36.9798 \approx 37 \ J$

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

The work done on the gas = 37 J.