IIT JEE past question Thermodynamics 3

JEE Main 2015 Guidance › Thermal physics questions › IIT JEE past question Thermodynamics 3

A monoatomic ideal gas of two moles is taken through a cyclic process starting for A as shown in the figure .The volume ratios are V_B/V_A=2 and V_D/V_A=4. If the temperature T_A at A is 27 degree C, calculate
a) The temperature of gas at point B
b) Heat absorved or released by the gas in each process
c) The total work done by the gas during the complete cycle.

Express answer in terms of R.

Diagram coming in a bit ;)

#Physics #Thermal physics

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7 Answers

62

Lokesh Verma ·2009-03-28 07:15:07

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tapanmast Vora ·2009-03-28 07:21:58

"VD/VC=4."

seems to b a typo ^^^

coz frm diag. Vc = Vd

Is it Td/Tc = 4 ??

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ANKIT MAHATO ·2009-03-28 07:22:33

giving !

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ANKIT MAHATO ·2009-03-28 07:23:38

using geometrical concept
Tb/Ta = Vb/ Va
Tb = 300 * 2 = 600K

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tapanmast Vora ·2009-03-28 07:25:43

a) 600K

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ANKIT MAHATO ·2009-03-28 07:37:37

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ANKIT MAHATO ·2009-03-28 07:38:27

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Your Answer

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