SSC JE Mechanical Engineering 2018 Set 3 Paper

Question : 146 of 200

Marks: +1, -0

A gas is compressed frictionless from an initial state of 0.4 m

^{3}

and 0.1 MPa to a final state of 0.2 m

^{3}

and 0.1 MPa. There is a transfer of 35 kJ of heat from the gas. What is the change in internal energy (in kJ)?

Solution:

Given

V_1 = 0.4 m^3 , V_2 = 0.2 m^3

P_-1 = 0.1 \mPa = P_2

at constant pressure workdone

w = P.dν = 0.1 × 10^6 [0.2 -0.4]

= – 20 KJ
Now dQ = dν + dw ⇒ – 35 = dν + (–20)
⇒ dν = - 15 KJ

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