One mole of a monatomic ideal gas is taken through the cycle shown in diagram A andrarr;B adiabatic expansion B andrarr;C cooling at constant volume C andrarr;D adiabatic compression D andrarr;A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A andrarr;B, the heat lost by the gas in the process B andrarr;C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)andDelta;Q = 4385.435 J; TD = 400 Kb)andDelta;Q = - 5297.625 J; TD = 500 Kc)andDelta;Q = 6532.365 J; TD = 620 Kd)andDelta;Q = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer?

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One mole of a monatomic ideal gas is taken through the cycle shown in diagram
A → B adiabatic expansion
B → C cooling at constant volume
C → D adiabatic compression
D → A heating at constant volume
The pressure and temperature at A, B, etc., are denoted by P_A, T_A; P_B, T_B, etc., respectively.
Given T_A = 1000 K, P_B = (2/3) P_A and P_C = (1/3) P_A.
Calculate the work done by the gas in the process A → B, the heat lost by the gas in the process B → C and temperature T_D.
Given (2/3)^2/5 = 0.85 and R = 8.31 J/mol K.

a)
ΔQ = 4385.435 J; TD = 400 K
b)
ΔQ = - 5297.625 J; TD = 500 K
c)
ΔQ = 6532.365 J; TD = 620 K
d)
ΔQ = - 7684.515 J; TD = 750 K

Correct answer is option 'B'. Can you explain this answer?

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One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer?

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One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer?.

Solutions for One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.

Here you can find the meaning of One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer?, a detailed solution for One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer? has been provided alongside types of One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer? theory, EduRev gives you an ample number of questions to practice One mole of a monatomic ideal gas is taken through the cycle shown in diagram A →B adiabatic expansion B →C cooling at constant volume C →D adiabatic compression D →A heating at constant volumeThe pressure and temperature at A, B, etc., are denoted by PA, TA; PB, TB, etc., respectively.Given TA = 1000 K, PB = (2/3) PA and PC = (1/3) PA.Calculate the work done by the gas in the process A →B, the heat lost by the gas in the process B →C and temperature TD.Given (2/3)2/5 = 0.85 and R = 8.31 J/mol K.a)ΔQ = 4385.435 J; TD = 400 Kb)ΔQ = - 5297.625 J; TD = 500 Kc)ΔQ = 6532.365 J; TD = 620 Kd)ΔQ = - 7684.515 J; TD = 750 KCorrect answer is option 'B'. Can you explain this answer? tests, examples and also practice JEE tests.

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