- Note
- Things to remember

Thermodynamics is the branch of science which deals with the measurement of transformation if heat into mechanical work. That means it mainly describes the inter-relationship between heat and mechanical energy.

**Thermodynamic System**

It is an assembly of an extremely large number of particles having a certain pressure, volume and temperature. A thermodynamic system can exchange energy with its surrounding by transfer or do mechanical work. The state of a system is the physical condition of system which are described by pressure, volume and temperature. A thermodynamic system may be isolated or closed system.

**Thermal Equilibrium**

A thermodynamic system is in thermal equilibrium if the temperature of all parts of it is same. So, a system in thermal equilibrium does not exchange heat between its different parts or with surroundings.

Let us consider an ideal gas is enclosed in a cylinder fitted with a frictionless moveable piston. Let the pressure exerted by the gas be P and the cross-sectional area of the piston be A. The force exerted by the gas on the piston is F= PA.

When the gas expands, the piston moves out through a small distance dx

and work done by the force

$$dW = F dx = PA dx$$

Since \(A dx = dV\), a small increase in the volume of the gas, the work done by the gas during the expansion.

$$dW = PdV \dots (i)$$

When the volume of the gas changes from V_{1} to V_{2,} the total work done W is obtained by integrating the equation (i) within the limits V_{1} to V_{2}.

$$W = \int dW = \int _{V_1}^{V_2} P dV \dots (ii)$$

When the final volume V_{2} is greater than initial volume V_{1}, then the change in volume V_{2}- V_{1} is positive. Hence during the expansion of the gas work done by a system is taken as positive.

When the gas is compressed, the final volume V_{2} is less than the initial volume V_{1}then the change in volume V_{2}- V_{1} is negative.Hence during the compression of the gas work done by a system is taken as negative.

Let E be a point on the indicator diagram. P and V be the pressure and volume at the point. Let the volume increase by small amount dV at constant P to a point F which is close to E.

From the figure, we have

Area of the small strip \( EFGH = HG \times EH = dV \times P = PdV \) where dV is a small volume represented by HG during which the pressure P = HE remains constant.

\(\therefore\) The area of the strip EFGH = PdV = work done during a small change of volume dV.

Thus, the work done during expansion from the initial state A(P_{1}, V_{1}) to the final state (P_{2}, V_{2}) can be obtained by adding the area of such small stripes

$$W = \text {area of ABCD}$$

Thus work done by a system is numerically equal to the area under PV-diagram. So,

$$W = \int _{V_1}^{V_2} P dV= \text {area of ABCD} $$

When the pressure P remains constant throughout the expansion, the work done by the gas is

$$W = P(V_2 - V_1)$$

After thecertain process, if the system returns to its initial state is cyclic process is called cyclic process.

Let us consider a system expand from Initial stage A to the state B along the path X and the system compress from B to A along the path Y which is shown in the figure.

The work done by system when system expand from A to B is represented by the curve AXB and volume axis that is

$$W_1 = \text {area of AXBB'A'A}$$

The work done on the system when system contract fromBA is given by the area between curve BYA and volume axis i.e.

$$W_2 = -(\text {area of BYAA'B'B})$$

$$\therefore \text {Net work done in the cyclic process is}$$

$$W = W_1 + W_2$$

$$= \text {area of AXBB'A'A} +(\text {-area of BYAA'B'B})$$

$$= + \text {area of AXBYA} $$

Thermodynamics is the branch of science which deals with the measurement of transformation if heat into mechanical work.

A thermodynamic system is in thermal equilibrium if the temperature of all parts of it is same.

After certain process if the system returns to its initial state is cyclic process is called cyclic process.

.-
## You scored /0

## ASK ANY QUESTION ON Thermodynamic System and Work Done

You must login to reply

## Nischal

Maths error

Mar 02, 2017

0 Replies

Successfully Posted ...

## Please Wait...