Hopeless MCQ....

Multiple Option correct
1 mole of a monoatomic gas undergoes a thermodynamic proces such that V^3=kT^2 where 'k' is a const.- and the other symbols bear the usual meanings.
1) Work done by the gas is 200R when temp is raised by 200K.
2) Specific Heat of the gas in the process is \frac{13R}{6}
3) When temp is raised by 600K, work done by the gas is 300R
4) Specific heat of the gas in the process is \frac{21R}{4}

5 Answers

1
decoder ·

is the answer (2)

11
Devil ·

God knows the ans.....plz give soln.

1
decoder ·

it's PV^{\frac{-1}{2}}=c

now W=\frac{nRdT}{1-\gamma}

where γ=-1/2

when dT=200K
W=400R/3

and when dT=600K
W=400R

and for specific heat capacity it's

C=C_{v} + \frac{PdV}{ndT}

just differentiate V^{3}T^{-2}=c w.rt T

and put dV/dT u will get the answer

1
varun.tinkle ·

remember this shortcut c=cV +c/1-x
here x is pv^x=k

11
Devil ·

Thanx all of u.

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