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Admission to master's degree courses in Delhi University is conducted in two ways, first on the basis of the merit of the qualifying examination which is specifically for students of delhi university and since maharaj agrasen college is a constituent college of delhi university, hence if you fulfill the
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Hello Dear Student,
For Admission in LLB Course, Eligibility Criteria is Graduation So you just need the Marksheets of your Graduation to participate in the Counselling process of LLB Course, If you have degree certificate of your Graduation too then it will be more better, you don't need to show
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T1 = 27 = 300K
Let initial pressure be P1 and final be P2 .
Now, as gas is compressed to 10 times the original pressure then :-
P1/P2 = 1/10
And, gamma = 1.4
And since the gas is compressed, so we will use
(T1 / T2 )^gamma =
Dear student,
T1 = 10 = 283K
Let initial volume be V1 and final be V2 .
Now, as gas is adiabatically compressed to one third of it's original volume then :
V2 = 1/3 of V1
That is :-
V2 / V1 = 1/3
And, gamma = 1.5
And
Hello there!
Here is the solution regarding your question.
Given,
T = 27°C [which is equal to 300 K].
Work done = 2.303 RT log Vf / Vi [ for isothermal process ].
W = 2.303 x 8.31 x 300 x log 3/1 [ log 3 = 0.48 ].
W
T = 127°C [which is equal to 400 K].
Pi = 1 x 10 dyne / cm²
Pf = 5 x 10 dyne / cm²
Since, the process is Isothermal. Thus,
PiVi = PfVf
Vf = PiVi / Pf
Vf
Vi = 1.2 litres
T = 37°C [ and that is equal to 310 K ].
Pi = 10 dyne / cm²
Vf = 3 x Vi
The process mentioned in the question is Isothermal. So,
PV = constant
PiVi=
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