Question

A stationary mass of gas is compressed without friction from an initial state of 0.5 m³ to and 0.2 MPa to a final state of 0.3 m³ and 0.2 MPa. Determine the work done by the system.
1. 40 kJ
2. 30 kJ
3. 50 kJ
4. 80 kJ

Answer 1

Correct Answer - Option 1 : 40 kJ

Concept:

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

For the Isobaric process, the work done is given by,

W = P ΔV = P(V1 - V2)

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

Calculation:

Given:

P = 0.2 MPa = 0.2 × 10⁶ N/m², V₁ = 0.5 m³, V₂ = 0.3 m³

W1-2 = P(V1 - V2)

= 0.2 × 10⁶(0.5 - 0.3)

= 40 kJ

 

Adiabatic process:

• It is a thermodynamic process in which there is no exchange of heat from the system to its surroundings 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 Cp and Cv 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 the Isothermal process work is done is given by, ​​\({W_{1 - 2}} = {p_1}{V_1}\ln \left( {\frac{{{p_1}}}{{{p_2}}}} \right) = {p_2}{V_2}\ln \left( {\frac{{{p_1}}}{{{p_2}}}} \right)\)