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During change in the state of a system, energy is transferred either towards (or) from the surroundings. This energy may be transferred as heat (or) mechanical work. In thermodynamics, the only type of work generally considered is the work done in expansion (or) compression of a gas. This is known as pressure volume work (or) PV work.

Thermodynamics is concerned with mechanical work, also known as pressure-volume work (or) expansion work.

Change in volume = Distance (l) x Area (A)

ΔV = V_{f} - V_{i
}Pressure = Force /Area

Force on Piston (F) = Pressure x Area

F = P_{ex} x A

Work(w) = Force (F) x Distance (l)

Force (F) = P_{ex} x A

On substituting the value of F in the equation for work we get...

w = P_{ex} x A x l

l x A = ΔV

w = P_{ex} x (-ΔV)

w = P_{ex} x (V_{f} - V_{i})

A reversible process is one that takes place infinitesimally slowly and the direction of which at any point can be reversed by an infinitesimal change in the state of the system. In a reversible process the system is in equilibrium in the initial, final and all intermediate stages. Processes, which are not reversible, are known as irreversible processes. Under reversible conditions, the relationship between work and internal pressure is written as reversible work,

w = - ∫_{vi}^{vf} P_{ex}dV

We know Pex = (P_{in} ± dp)

On replacing we get

W_{rev} = - ∫_{vi}^{vf} (P_{in} ± dp) dV

W_{rev} = - ∫_{vi}^{vf} P_{in} dV

pV = nRT or p = nRT/V

W_{rev} = - ∫_{vi}^{vf} nRT/V dV

= - nRT ln(v_{f}/v_{i})

W_{rev} = -2.303nRTlogV_{f}/V_{i}

If the gas expands in vacuum, external pressure P_{ex} being zero in the expression

W =-Pex.∆V

W = 0

This is known as free expansion. No work is done during the free expansion of an ideal gas whether the process is reversible (or) irreversible.

According to the first law of thermodynamics the internal energy is,

∆U =q + w

On substituting the value of w in the equation,

∆U = q + (- Pex.∆V)

If the process is carried out at constant volume, then ∆V is zero. Therefore work done is also zero.

No work is done during free expansion and change in internal energy,

∆U =q_{v,} where q_{v} is heat supplied at constant volume.

During change in the state of a system, energy is transferred either towards (or) from the surroundings. This energy may be transferred as heat (or) mechanical work. In thermodynamics, the only type of work generally considered is the work done in expansion (or) compression of a gas. This is known as pressure volume work (or) PV work.

Thermodynamics is concerned with mechanical work, also known as pressure-volume work (or) expansion work.

Change in volume = Distance (l) x Area (A)

ΔV = V_{f} - V_{i
}Pressure = Force /Area

Force on Piston (F) = Pressure x Area

F = P_{ex} x A

Work(w) = Force (F) x Distance (l)

Force (F) = P_{ex} x A

On substituting the value of F in the equation for work we get...

w = P_{ex} x A x l

l x A = ΔV

w = P_{ex} x (-ΔV)

w = P_{ex} x (V_{f} - V_{i})

A reversible process is one that takes place infinitesimally slowly and the direction of which at any point can be reversed by an infinitesimal change in the state of the system. In a reversible process the system is in equilibrium in the initial, final and all intermediate stages. Processes, which are not reversible, are known as irreversible processes. Under reversible conditions, the relationship between work and internal pressure is written as reversible work,

w = - ∫_{vi}^{vf} P_{ex}dV

We know Pex = (P_{in} ± dp)

On replacing we get

W_{rev} = - ∫_{vi}^{vf} (P_{in} ± dp) dV

W_{rev} = - ∫_{vi}^{vf} P_{in} dV

pV = nRT or p = nRT/V

W_{rev} = - ∫_{vi}^{vf} nRT/V dV

= - nRT ln(v_{f}/v_{i})

W_{rev} = -2.303nRTlogV_{f}/V_{i}

If the gas expands in vacuum, external pressure P_{ex} being zero in the expression

W =-Pex.∆V

W = 0

This is known as free expansion. No work is done during the free expansion of an ideal gas whether the process is reversible (or) irreversible.

According to the first law of thermodynamics the internal energy is,

∆U =q + w

On substituting the value of w in the equation,

∆U = q + (- Pex.∆V)

If the process is carried out at constant volume, then ∆V is zero. Therefore work done is also zero.

No work is done during free expansion and change in internal energy,

∆U =q_{v,} where q_{v} is heat supplied at constant volume.