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A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared
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the JEE exam syllabus. Information about A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2024 Exam.
Find important definitions, questions, meanings, examples, exercises and tests below for A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer?.
Solutions for A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer? in English & in Hindi are available as part of our courses for JEE.
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Here you can find the meaning of A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer? defined & explained in the simplest way possible. Besides giving the explanation of
A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer?, a detailed solution for A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer? has been provided alongside types of A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
ample number of questions to practice A gaseous sample is generally allowed to do only expansion/compression type of work against its surroundings. The work done in case of an irreversible expansion (in the intermediate stages, of expansion/compression the states of gases are not defined). The work done can be calculated usingdW = NextdVwhile in case of reversible process the work done can be calculated using dw = PdV where P is pressure of gas at some intermediate stages. Like for an isothermal reversible process.Since d w = P d V so magnitude of worked one can also be calculated by, calculating the area under the PV curve of the reversible process in PV diagram.Q.An ideal gaseous sample at initial state i (P0V0T0) is allowed to expand to volume 2V0 using two different process; in the first process the equation of process is PV2= K1 and in second process the equation of the process is PV = K2 .Thena)work done in first process will be greater than work in second process (magnitude wise)b)The order of values of work done can not be compared unless we know the value of K1 and K2.c)value of work done (magnitude) in second process is greater in above expansion irrespective of the value of K1 and K2 .d)Ist process is not possibleCorrect answer is option 'C'. Can you explain this answer? tests, examples and also practice JEE tests.