11ΔV = k(t2 + 3600 - 120T)
W = pΔV
W =pk(t2 + 3600 - 120T)
W is the work done by the gas
1 mole of a gas xpands with temperature T such dat its volume V=kT2 where k is a constant if d temperature of d gas changs by 60° den find d work done by d gas ?
a gas has molar heat capacity C=37.35Jmol^-1 K^-1 in d process PT=k find d no of degrees of freedom of d molecules in d gas
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9 Answers
11
1V=KT2
T/P=KT2
P=R/(KT)
W=∫P.dv
dv=2TK
W=∫1/(KT)2KTdT
W=2*(T2-T)
W=2*60R
W=120R
21that to is crrct only,
i wanna know in ram's methd wer ddid R cum frm?
1INSTEAD OF nRdeltaT
HE FORGOT WRITIN R in one step
but in final he didnt !
W=2*(T2-T)R
W=2*60R