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GATE previous years thermodynamics solved questions
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In this video we will solve some GATE previous year thermodynamics questions .

Rutuja Bothara
Strengthen GATE /PSU/ESE/SSC JE/RRB JE /campus Placements preparation. A passionate mechanical engineer -VIT University

Unacademy user
mam for ideal gas joule Thompson coefficient is zero... so delta T should be zero. Am I right or wrong?
Rutuja Bothara
6 months ago
Yes. You are right. Since for an ideal gas the entrance is function of temperature only, at constant enthalpy process temperature remains constant. So delta T =0
Rutuja Bothara
6 months ago
Vishal tewari
6 months ago
thanks 😊
Rutuja Bothara
6 months ago
welcome :)
  1. GATE Previous years Thermodynamics questions Important Two-marks GATE questions 7th lesson By Rutuja Bothara

  2. RUTUJA BOTHARA . Perusing B.Tech in Mechanical Engineering from Vellore Institute of Technology, Vellore. (2015-2019) Cleared JEE Mains 2015 Cleared VITEEE-2015 with AIR 8718 . Cracked HITSEEE-2015 with AIR 91 Worked as a Career Counsellor in Teachers Academy . (2016- 2018) and BOLO app.

  3. Nitrogen gas (molecular weight 28) is enclosed in a cylinder by a piston, at the initial condition of 2 bar, 298 K and 1. In a process, the gas slowly expands under isothermal condition, until the volume becomes 2 Heat exchange occurs with the atmosphere at 298 K during this process. The work interaction for the nitrogen gas is #200 Kj 138.6 KJ - -200 K

  4. Work done in isothaymal process VI 2x105* In 138G KJ

  5. .The entropy change for universe during the process in KJ/K is 0.4652 0.0067 0.0 -0.06711

  6. Fr hd la of ue modynamics, tor2 a n ICS entrory chan is ien by K 0.4652 |

  7. A Carnot cycle is having an efficiency of 0.75. If the temperature of the high temperature reservoir is 727 C, what is the temperature of the low temperature reservoir? 23 C -23 C 0 C 250 C

  8. Given High tempetatre 1 2

  9. A cyclic heat engine does 50 KJ of work per cycle. If the efficiency of the heat engine is 75 %, the heat rejected per cycle is -16 (2/3) 33(1/3) #37(1/2) -66(2/3)

  10. Essiciene of heat engin work done Heat taken Heat gien-work done Heat ejectel Heat ejecte dl 3

  11. steady Flouo ehergs equation is 2 2 2. adiabatic proces udith hegligible changeh in kE and P. E hi h2 s >0 , because of irreversible proc et

  12. Given V1=0, According to congervatHon of ener 2 2 2 hi-h2)2 | 0.8 2x1D