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A movable piston traps 0.205 moles of an ideal gas in a vertical cylinder. If the piston slides without friction in the cylinder, how much work will be done on the gas when its temperature is decreased from 270 ∘C to 24∘C? Hint: Since the piston slides without friction, the pressure will remain constant. Use the ideal gas law to change W=−PΔ???? to an expression using ???? and T. Be sure to include the appropriate sign of the work in the answer.

Respuesta :

Answer : The work done on the gas will be, 418.4 J

Explanation :

First we have to calculate the volume at 270°C.

[tex]PV_1=nRT[/tex]

where,

P = pressure of gas = 1 atm

[tex]V_1[/tex] = volume of gas = ?

T = temperature of gas = [tex]270^oC=273+270=543K[/tex]

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

[tex](1atm)\times V_1=0.205mol\times 0.0821L.atm/mol.K\times 543K[/tex]

[tex]V_1=9.12L[/tex]

Now we have to calculate the volume at 24°C.

[tex]PV_2=nRT[/tex]

where,

P = pressure of gas = 1 atm

[tex]V_2[/tex] = volume of gas = ?

T = temperature of gas = [tex]24^oC=273+24=297K[/tex]

n = number of moles of gas = 0.205 mol

R = gas constant  = 0.0821 L.atm/mol.K

Now put all the given values in the ideal gas equation, we get:

[tex](1atm)\times V_2=0.205mol\times 0.0821L.atm/mol.K\times 297K[/tex]

[tex]V_2=4.99L[/tex]

Now we have to calculate the work done.

Formula used :

[tex]w=-p\Delta V\\\\w=-p(V_2-V_1)[/tex]

where,

w = work done

p = pressure of the gas = 1 atm

[tex]V_1[/tex] = initial volume = 9.12 L

[tex]V_2[/tex] = final volume = 4.99 L

Now put all the given values in the above formula, we get:

[tex]w=-p(V_2-V_1)[/tex]

[tex]w=-(1atm)\times (4.99-9.12)L[/tex]

[tex]w=4.13L.artm=4.13\times 101.3J=418.4J[/tex]

conversion used : (1 L.atm = 101.3 J)

Therefore, the work done on the gas will be, 418.4 J

The work done in the constant pressure system of the gas is given by the

product of the change in volume and gas pressure.

Work done on the gas is approximately 419.3 Joules

Reasons:

The given parameters are;

Number of moles of gas, n = 0.205 moles

Type of gas = Ideal gas

Initial temperature of the gas, T₁ = 270 °C

Final temperature of the gas, T₂ = 24 °C

Required:

Amount of work done on the gas

Solution:

Work done on the gas, W = -P·ΔV

The ideal gas equation is; P·V = n·R·T

P·V₁ = n·R·T₁

P·V₂ = n·R·T₂

P·ΔV = P·(V₂ - V₁) = P·V₂ - P·V₁

∴ Work done of the gas, W =  -P·ΔV = -(n·R·T₂ - n·R·T₁) = -n·R·(T₂ - T₁)

The universal gas constant, R = 8.3145 J·mol⁻¹·K⁻¹

-P·ΔV = -(0.205 × 8.3145 × (24 - 270)) ≈ -(-419.3)

Work done on the gas, W ≈ 419.3 J

Learn more here:

https://brainly.com/question/14581707

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