Question

In which process the work-done is zero?
1. Adiabatic
2. Isothermal
3. Isochoric
4. Isobaric

Answer 1

Correct Answer - Option 3 : Isochoric

Explanation:

Work

• Work is said to be done by the system if the sole effect of the thing external to the system crosses the boundary.
• If there is no interaction of work between system and surrounding then work done is zero.
• Work is high-grade energy.
• It is also known as an organized form of energy.

Isochoric process:

• It is a thermodynamic process which takes place at constant volume.
• In such a process, the work done is zero. The volume of the gas remains constant.

For constant volume or Isochoric process work done is given by,

\(W = \smallint PdV\)

Since for isochoric process we know that,

dV = 0

so, W = 0 for Isochoric process.

1. For free expansion work done is zero. i.e. W = 0

2. The process in which high-pressure fluid is converted to low pressure by using a throttle valve is Throttling. In the throttling process enthalpy remains constant, hence work done is zero.

Adiabatic process:

• It is a thermodynamic process in which there is no exchange of heat from the system to its surrounding neither during expansion nor during compression, yet the temperature may change.
• For the adiabatic process to occur, the system must be perfectly insulated from the surrounding.
• A reversible adiabatic process is called isentropic.

For the adiabatic process, the work done by the system is given as

\(W = \frac{{{P_1}{V_1} - {P_2}{V_2}}}{{\gamma - 1}} = \frac{{nR\left( {{T_1} - {T_2}} \right)}}{{\gamma - 1}}\)

where R is gas constant, n is molar mass and \(\gamma = \frac{{{C_P}}}{{{C_V}}}\) , where C_p and C_v are specific heat at constant pressure and volume

Isothermal process

• It is a thermodynamic process in which the temperature of a system remains constant.
• The thermal equilibrium is maintained because the heat transfer is very slow.
• The work done in this process is due to the change of net heat content of the system.

For Isothermal process work done is given by,

\({W_{1 - 2}} = {p_1}{V_1}\ln \left( {\frac{{{p_1}}}{{{p_2}}}} \right) = {p_2}{V_2}\ln \left( {\frac{{{p_1}}}{{{p_2}}}} \right)\)

Isobaric process:

• It is a thermodynamic process in which the pressure remains constant.

For Isobaric process the work done is given by,

W = P ΔV 

where P = pressure in N/m², ΔV = Change in volume.