GATE Mechanical Engineering (ME) 2020 Shift 2 Solved Paper

Question : 45 of 65

Marks: +1, -0

Air is contained in a frictionless piston-cylinder arrangement as shown in the figure.

The atmospheric pressure is

100 {kPa}

and the initial pressure of air in the cylinder is 105 kPa. The area of piston is

300 {cm}^{2}

. Heat is now added and the piston moves slowly from its initial position until it reaches the stops. The spring constant of the linear spring is

12.5

N/mm. Considering the air inside the cylinder as the system, the work interaction is ______________ J (round off to the nearest integer).

Your Answer:

Validate

Solution:

Considering Air inside the cylinder as the system
Total work by gas = work done by the spring + work done by weight of piston + Work done by atmosphere
Work done by spring

=\;{1}/{2} {k}( {x}_{2}^{2}- {x}_{1}^{2})

=\;{1}/{2} × 12.5 × 10^{3} ×(0.08)^{2}

=40 {J}

Work done by atmosphere

=

Patm

\times

volume

=100 × 10^{3} × 0.03 × 0.16

=480 {J}

Work done by weight of piston:
At initial state

P_{\;\text "initial "\;} ×

Area

=P_{\;\text "atm "\;} ×

Area

+W_{\;\text "piston "\;}

105 × 10^{3}=100 × 10^{3}+\;{W_{\;\text "piston "\;}}/{0.03}

W_{\;\text "piston "\;}=150 {N}

Work done by piston

(W_{\;\text "piston "\;})=150 × 0.16=24

Joule
So Total work done by gas

=40+480+24

=544

Joule

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