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Calculate the max. work done by system in an irreversible (single step) adiabatic expansion of 2 mole of a polyatomic gas from 300K and pressure 10 atm to 1 atm.( γ = 1.33).?
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Calculate the max. work done by system in an irreversible (single step...
Max Work Done by System in Irreversible Adiabatic Expansion


Given:


  • Number of moles of gas (n) = 2

  • Initial temperature (T1) = 300 K

  • Initial pressure (P1) = 10 atm

  • Final pressure (P2) = 1 atm

  • Ratio of specific heats (γ) = 1.33



Calculating the Max Work Done:


Step 1: Determining the Initial Volume (V1):

We can use the ideal gas law to find the initial volume of the gas:
PV = nRT

Substituting the given values:
V1 = (nRT1) / P1
V1 = (2 * 0.0821 * 300) / 10
V1 = 4.923 L

Step 2: Determining the Final Volume (V2):

Since the process is irreversible, we cannot directly use the adiabatic equation P1V1^γ = P2V2^γ. Instead, we use the relationship between temperature and volume for an adiabatic process:
T1V1^(γ-1) = T2V2^(γ-1)

Substituting the given values:
300 * (4.923)^(1.33-1) = T2 * V2^(1.33-1)
T2 * V2^0.33 = 300 * 4.923^0.33
V2^0.33 = (300 * 4.923^0.33) / T2

Step 3: Applying the Ideal Gas Law to Find Final Volume:

Using the ideal gas law, we can express the final volume in terms of pressure and temperature:
PV = nRT

Substituting the given values:
1 * V2 = 2 * 0.0821 * T2
V2 = (2 * 0.0821 * T2) / 1

Step 4: Combining Equations to Solve for Final Volume:

Substituting the expression for V2 from Step 3 into the equation from Step 2:
((2 * 0.0821 * T2) / 1)^0.33 = (300 * 4.923^0.33) / T2

Simplifying the equation:
(2 * 0.0821 * T2) / 1 = (300 * 4.923^0.33) / T2^(0.33)
T2^1.33 = (300 * 4.923^0.33) / (2 * 0.0821)
T2 = [(300 * 4.923^0.33) / (2 * 0.0821)]^(1/1.33)
T2 ≈ 468.3 K

V2 = (2 * 0.0821 * 468.3
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A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not define d). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.If four identical samples of an ideal gas initially at similar state (P0, Vo, T0) are allowed to expand to double their volumes by four different processes..I : by isothermal irreversible process II : by reversible process having equation P2V = constant III : by reversible adiabatic process IV: by irreversible adiabatic expansion against constant external pressure.Then in the graph shown the final state is represented by four different points then, the correct match can be

A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not define d). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.Two samples (initially under same states) of an ideal gas are first allowed to expand to double their volume using irreversible isothermal expansion against constant external pressure, then samples are returned back to their original volume first by reversible adiabatic process and second by reversible process having equation PV2 = constant then

A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not define d). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Then

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Calculate the max. work done by system in an irreversible (single step) adiabatic expansion of 2 mole of a polyatomic gas from 300K and pressure 10 atm to 1 atm.( γ = 1.33).?
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