Respuesta :
Answer:
The volume needed to expand the gas is = 11 Lit = 11000 [tex]cm^{3}[/tex]
Explanation:
Given :
Initial volume [tex]V_{1} = 40 cm^{3} = 0.04[/tex] Lit.
Energy [tex]W = 950[/tex] J = 9.38 Lit × atm. ⇒ ( 1 Lit×atm. = 101.325 J )
Pressure [tex]P =[/tex] 650 torr = 0.855 atm. ⇒ ( 1 torr = 0.00132 atm )
In this example we have to be aware of unit conversion system.
From the laws of thermodynamics,
[tex]W = P \Delta V[/tex]
Here in this example, all the energy of combustion is converted into work to push back the piston
[tex]W = P (V_{2} - V_{1} )[/tex]
[tex]V_{2} - 0.04 = 10.96[/tex]
[tex]V_{2} = 11[/tex] Lit = [tex]11000 cm^{3}[/tex]
The given system is a constant pressure system where the work done is the product of the pressure and the volume change.
- The volume to which the gas expands to is approximately 11 liters
Reasons:
Given parameters;
The original volume, V₁ = 40 cm³
The energy released, E = 950 J
Constant pressure applied, P = 650 torr.
Condition: All energy is converted to work to push back the piston
Required:
The volume to which the gas will expand
Solution;
The work done, W = Energy releases, E = 950 J
Work done at constant pressure, W = P·(V₂ - V₁)
Where;
V₂ - The volume to which the gas will expand
Converting the volume to from cm³ to m³ gives;
V₁ = 40 cm³ = 0.00004 m³
Converting the pressure given in torr to Pascals gives;
650 torr. = 86659.54 Pa.
Therefore, we get;
950 J = 86659.54 Pa. × (V₂ - 0.00004 m³)
[tex]V_2 = \dfrac{950 \, J}{86659.54 \, Pa} + 0.00004 \, m^3 = \dfrac{2888}{263445} \ m^3 + 0.00004 \, m^3 = \dfrac{2683}{243855} \, m^3[/tex]
Converting to liters gives;
[tex]V_2 =\dfrac{2683}{243855} \, m^3 \times \dfrac{1,000 \, l }{m^3} \approx 11.00244\, l \approx 11 \, l[/tex]
- The volume to which the gas expands, V₂ ≈ 11 liters.
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