Application of the first law of thermodynamics to various closed systems

First law of thermodynamics to Reversible Constant Volume Process

In a constant volume process ( v = constant), the working substance is contained in a rigid vessel, So the boundary of the system is being rigid and no work interaction between system and surroundings other than paddle wheel work input. It will assume that “Constant volume” implies zero work unless stated otherwise.

(Fig. shows System and the states before and after heat addition at constant Volume)

Considering the mass of the working substance unity and applying the first law of thermodynamics into the process

$Q = \left( u_2 - u_1 \right) + W$

The work done $W = \int_1^2 pdv = 0$ as $dv = 0$

$\therefore \boxed{Q = \left( u_2 - u_1 \right) = c_v \left( T_2 - T_1 \right) }$

where $c_v$ is the specific heat at constant volume.

For mass m of working substance :

$\boxed{Q = U_2 - U_1 = m. c_v. \left(T_2 - T_1 \right)}$

Since, $m.u =U$

The first law of thermodynamics to Reversible Constant Pressure Process

In the constant volume process, the boundary of the system is rigid and the pressure rises when the heat is supplied. So for the constant pressure process, the boundary must move against an external resistance when the heat is supplied. For example, A gas behind the piston in the cylinder may undergo a constant pressure process. Since the piston is pushed by a certain distance by the force produced by the gas, then the work is done by the gas to the surroundings.

Considering the unit mass of working substance and applying the first law of thermodynamics to the process.

$Q = \left( u_2 - u_1 \right) + W$

The work done, $W = \int_1^2 p.dv = p \left( v_2 - v_1 \right)$

$\therefore Q = \left( u_2 - u_1 \right) + p \left( v_2 - v_1 \right)$

$\Rightarrow Q = \left( u_2 - u_1 \right) + p . v_2 - p. v_1$

$\Rightarrow Q = \left( u_2 + p. u_2 \right) - \left( u_1 + p . v_1 \right)$

$= h_2 - h_1$

or $\boxed{ Q = h_2 - h_1 = c_p \left( T_2 -T_1 \right) }$

where, h = Enthaly ( specific)

c_p = Specific heat at constant pressure

For mass, m of working substance:

$\boxed{ Q = \left( H_2 - H_1 \right) = m. c_p. \left(T_2 - T_1 \right)}$

First law of thermodynamics to Reversible Isothermal Process

“A process at a constant temperature is called an Isothermal Process ( p.v = Constant and T = Constant).” In an isothermal expansion of a working substance in a cylinder behind the piston expands from high pressure to a low pressure, there is a tendency to fall the temperature. In an isothermal expansion heat must be added continuously to keep temperature constant. Like wise in an isothermal compression , there is a tendency to temperature rise so heat must be removed from working substance during the process.

Considering the unit mass of a working substance, By applying The First Law of Thermodynamics :

$Q = \left( u_2 - u_1 \right) + W$

$Q = c_v \left( T_2 - T_1 \right) + W$

$Q = 0 + W$

So, The work done , $W = \int_1^2p.dv$

In this case, p.v = Constant, so $p= \frac{C}{v}$

$\therefore W = \int_{v_{1}}^{v_{2}} C. \frac{dv}{v} = C. \left [ \log_e v \right ]_{v_1}^{v_2}$

$\Rightaroow W = C . \log_e . \frac{v_2}{v_1}$

The constant C can be written as $p_1 . v_1$ or $p_2 . v_2$

Since, $p_1 . v_1 = p_2 . v_2 = Constant, C$

So, $W = p_1 . v_1 \log_e \frac{v_2}{v_1}$ per unit mass

and

$W = p_2 . v_2 \log_e \frac{v_2}{v_1}$ per unit mass

$\therefore Q = W = p_1 . v_1 \log_e \frac{v_2}{v_1}$ ….(1)

for mass m, of the working substance

$\boxed{Q = p_1 . V_1 \log_e \frac{V_2}{V_1} }$

$\boxed{ Q = p_1 . V_1 \log_e \frac{p_1}{p_2}}$ $\because \frac{V_2}{V_1} = \frac{p_1}{p_2}$

First law of thermodynamics to Reversible Adiabatic Process

“An Adiabatic Process is one in which no heat is transferred to or from the fluid during the process. Such a process can be reversible or irreversible.”

The reversible adiabatic non-flow process will be considered in this section. Consider the unit mass of the working substance.

By applying the first Law of thermodynamics:

$Q = \left( u_2 - u_1 \right) + W$

Since, Q = 0 ( An Adiabatic Process)

So, $0 = \left( u_2 - u_1 \right) + W$

$\Rightarrow W = - \left( u_2 - u_1 \right)$

$\Rightarrow \boxed{W = \left( u_1 - u_2 \right)}$ …(1)

This equation is true for an adiabatic process whether the process reversible or irreversible process. In an adiabatic expansion, the work-done W by the fluid is at the expense of a reduction in the internal energy of the fluid. Similarly in an adiabatic compression process all the work done on the fluid goes to increase the internal energy of the fluid.

There must be perfect thermal insulation for the system, for an adiabatic process.

The pressure (p) and volume (v) relation for an adiabatic reversible process

To obtain a law relating to p and v for a reversible adiabatic process.

The non-flow energy equation in differential form:

$dQ = du + dW$

For a reversible process: $dW = p.dv$

$\therefore dQ = du + p.dv =0$ [since, for an adiabatic process Q = 0]

Also, For prefect gas, We know : $p.v =R.T \Rightarrow p = \frac{R.T}{v}$

So, $dQ = du + \frac{R.T.dv}{v} = 0$

also, $u = c_v. T \Rightarrow du = c_v. dT$

$\therefore c_v.dT + \frac{R.T.dv}{v} = 0$

Dividing both sides by T, we get

$c_v. \frac{dT}{T} + \frac{R.dv}{v} =0$

Integrating

$\Rightarrow c_v \log_e T + R. \log_e v = constant$

$\Rightarrow c_v \log_e \frac{p.v}{R} + R. \log_e v = constant$ ( $\because T = \frac{p.v}{R}$ )

Dividing bothsides with c_v , we have

$\log_e \frac{p.v}{R} + \frac{R}{c_v} . \log_e v = Constant$

we also know $c_v = \frac{R}{ \gamma -1}$ or $\frac{R}{c_v} = \gamma -1$

So,

$\log_e \frac{p.v}{R} + \left(\gamma - 1 \right) \log_e v =constant$

$\Rightarrow \log_e \frac{p.v}{R} + \log_e v^{\gamma -1} =constant$

$\Rightarrow \log_e \frac{p.v. \times v^{\gamma -1}}{R} = constant$

$\Rightarrow \log_e \frac{p.v^\gamma}{R} = constant$

$\Rightarrow \frac{p.v^\gamma}{R} = e^constant = constant$

or $\boxed{ p. v^\gamma = Constant}$

Equation of Work for adiabatic reversible process

A p-v diagram for a adiabatic reversible process is shown in fig.

The work-done is given by the green area and this area can be calculated by integration.

so,

$W = \int_{v_1}^{v_2} p. dv$

$\because p.v^\gamma = constant$

$\therefore W = \int_{v_1}^{v_2} C . \frac{dv}{v^\gamma}$ ( $\because p = \frac{dv}{v_\gamma}$ )

i.e $W = C \int_{v_1}^{v_2} \frac{dv}{v^\gamma}$

$\Rightarrow W = C \left | \frac{v^{-\gamma + 1}}{-\gamma + 1} \right |_{v_1}^{v_2}$

$\Rightarrow W = C \left( \frac{v_2^{- \gamma + 1} - v_1^{- \gamma + 1}}{1 - \gamma} \right)$

$\Rightarrow W = C \left( \frac{v_1^{- \gamma + 1} - v_2^{- \gamma + 1}}{ \gamma -1} \right)$

The constant in this equation can be written as $p_1 . v_1^\gamma$ or as $p_1 . v_1^\gamma$

Hence,

$W = \frac{p_1 . v_1 . v_1^{-\gamma + 1} - p_2. v_2. v_2^{-\gamma + 1}}{\gamma - 1}$

$\boxed{ W = \frac{p_1. v_1 - p_2. v_2}{\gamma -1}}$

Or $\boxed{ W = \frac{R \left(T_1 - T2 \right)}{\gamma -1}}$

Expression between Temperature (T) & Volume (v) and Temperature (T) & Pressure (p)

By using equation p.v = R.T

We have $p = \frac{R.T}{v}$

Putting the value of p = R.T /v in equation $p.v^\gamma = constant$

now we have, $\frac{R.T}{v} . v^\gamma = constant$

$\Rightarrow T. v^{\gamma - 1} = Constant$ . . . (1)

Also $v = \frac{R.T}{p}$ ; so again putting the value of v in equation $p.v^\gamma = constant$

now we have, $p. \left(\frac{R.T}{p}\right) ^\gamma = constant$

$\therefore \frac{T^\gamma}{p^{\gamma - 1}} = constant$

or $\frac{T}{p{\frac{\gamma - 1}{\gamma}}} = constant$ . . . (2)

Therefore, for a reversible adiabatic process for a perfect gas between states 1 and 2,

We can write like from equation $p. v^\gamma = constant$ ,

$\Rightarrow p_1 . v_1^\gamma = p_2 . v_2^\gamma$

$\Rightarrow \frac{p_2}{p_1} = \left( \frac{v_1}{v_2}\right)^\gamma$

From equation (1),

$\Rightarrow T_1 . v_1^{\gamma -1} = T_2 . v_2^{\gamma -1}$

$\Rightarrow \boxed{ \frac{T_2}{T_1} = \left( \frac{v_1}{v_2} \right)^{\gamma -1}}$

and from equation (2),

$\Rightarrow \frac{T_1}{p_1{\frac{\gamma - 1}{\gamma}}} = \frac{T_2}{p_2{\frac{\gamma - 1}{\gamma}}}$

$\Rightarrow \boxed{ \frac{T_2}{T_1} = \left( \frac{p_2}{p_1}\right)^{\frac{\gamma -1}{\gamma}}}$

Application of the first law of thermodynamics to various closed systems

First law of thermodynamics to Reversible Constant Volume Process

The first law of thermodynamics to Reversible Constant Pressure Process

First law of thermodynamics to Reversible Isothermal Process

First law of thermodynamics to Reversible Adiabatic Process

The pressure (p) and volume (v) relation for an adiabatic reversible process

Equation of Work for adiabatic reversible process

Expression between Temperature (T) & Volume (v) and Temperature (T) & Pressure (p)

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