Respuesta :
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;
[tex]\dfrac{p_1}{p_2} = \left [\dfrac{V}{v_1} \right ]^\gamma[/tex]
γ = 1.66 for a monoatomic gas, which gives;
[tex]\dfrac{1 \ atm}{0.5 \ atm} = \left [\dfrac{V}{1 \ liter} \right ]^{1.66}[/tex]
[tex]V = 1 \ liter \times \sqrt[1.66]{\dfrac{1 \ atm}{0.5 \ atm}} = 1 \ liter \times \sqrt[1.66]{2} \approx 1.52 \ liters[/tex]
V ≈ 1.52 liters = 0.00152 m³
The total work done is given given by the following relation;
[tex]W = \dfrac{K \times \left ( v_f^{1-\gamma}-v_i^{1-\gamma} \right )}{1 - \gamma}[/tex]
[tex]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}[/tex]
[tex]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[/tex]
Given that the work done is positive, we have that work is done in the gas
The work done on the gas = 37 J.
