One mole of an ideal gas expands from state-I (Atm , 20 litre) to (2 Atm , 10 litre ) isothermally .
Caluclate , `w & DeltaU`.
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A `27^(@)` one mole of an ideal gas is compressed isothermally and reversibly from a pressure of 2 atm to 10 atm , the value of `DeltaE and q` are (R
Correct Answer - a lsothermaly (At constant temperature) and reversibly work is ` W= 2.303 n RTlog(P_(2)/(P_(1)))= 2.303 xx 1xx2 xx 300 log(10/2)` 965.84 at constant temperature , `DeltaE=0` `DeltaE-q+ W,q--W=-965.84...
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One mole of an ideal gas is expanded isothermally at 300 K from 10 atm to 1 atm. Calculate q,w, DeltaU & DeltaH under the following conditions. (i) Ex
Isothermal process (i) For ideal gas `DeltaU=0" "DeltaH=0` `q= -w` `W_(rev) = - nRT ln ((P_(1))/(P_(r))) = - 1xxRxx300 In (10)/(1) = -690.9R` (ii) `W_(irrev) =- P_(ext)(V_(2)-V_(1)) = - P_(ext)((nRT)/(P_(2))-(nRT)/(P_(1)))...
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One mol an ideal gas is expanded from`(10 atm, 10lit)`. `(2 atm , 50 lit)` isothermally . First against 5 atm then against 2 atm . Calculate work done
`P_(1)" "V_(2) " "overset(Isothermal)to" "P_(1)" "V_(1)` 10 atm 10 lit `" "`2 atm 50 lit. (i) Work done against 5 atm pressure `(underset("10 atm")(P_(1))" "underset("10 lit")(V_(1)) to underset("5 atm ")(P_(2))underset("...
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One mole of an ideal gas is expanded isothermally jat `300 K ` from 10 atm to 1 atm . Find the values of `DeltaS_(sys),DeltaS_(surr),& DeltaS_("total"
(i) `DeltaS_("sys")= nC_(p) ln.(T_(2))/(T_(2)) + nR ln .(P_(1))/(P_(2))= 0 + R ln 10 = R ln 10` `DeltaS_("surr")= - DeltaS_("sys")= -R ln 10` `DeltaS_("total")= 0` (ii) ` DeltaS_("sys")= R ln...
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Calculate the change in entropy when 1 mole nitrogen gas expands isothermally and reversibly from an initial volume of 1 litre to a final volume of 10
`DeltaS=2.303nR" "log((V_(2))/(V_(1)))` `=2.303xx1xx8.314log((10)/(1))` `=19.12JK^(-1)`.
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(a) One mole of an ideal gas expands isothermally and reversibly at `25^(@)C` from a volume of `10` litres to a volume of `20` litres. (i) What is the
(a) (i) `DeltaS= 2.303 nR "log"(V_(2))/(V_(1))=2.303xx1xx8.314xx"log"(20)/(10)=5.76J//K`. (ii) `W_(rev)= -2.303 nRT "log"(V_(2))/(V_(1))` `= -2.303 xx1xx8.314xx298xx"log"(20)/(10)= -1781 J`. (iii) For isothermal process, `DeltaU=0` and heat is absorbed by the gas, `q_(rev)=DeltaU-W=0-(-1718)=1718J` `:....
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One mole of an ideal monoatomic gas expands isothermally against constant external pressure of `1 atm` from intial volume of `1L` to a state where its
Correct Answer - B `DeltaS =nR l n((V_(f))/(V_(i)))=R l n((P_(i))/(P_(f)))= R l n((300R)/(1Lxx1 atm))=R l n(24.6)`
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Calculate work done by `1` mole of ideal gas expand isothermally and irreversibly from pressure of `5` atm to `2` atm against a constant external pres
`int dW = -int P_("ext")dv` `W_("irr") = -P_("ext")[V_(2)-V_(1)] = -P_("ext")((nRT)/(P_(2))-(nRT)/(P_(1))) = -P_("ext") xx nRT((1)/(P_(2))-(1)/(1P_(1)))` `= -1 xx 1 xx .082 xx 300 ((1)/(2)-(1)/(5)) = -1 xx .082 xx 300 xx...
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If 1.5 L of an ideal gas at a apressure of 20 atm expands isothermally and reversibly to a final volume of 15 L, the work done by the gas in L atm is:
Correct Answer - D `w_("expansion")=-2.303nRT"log"_(10)((V_(2))/(V_(1)))` `=-2.303PV" "log_(10)((V_(2))/(V_(1)))` `=-2.303xx1.5xx20" log "((15)/(1.5))` `=69.09L" "atm`
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`10` litre of a monoatomic ideal gas at `0^(@)C` and `10 atm` pressure is suddenly released to `1 atm` pressure and the gas expands adiabatically agai
Correct Answer - 64 L This is adiabatic irreversible process, so for this process `PV^(gamma)=` Constant, is not applicable `W= -P_(ext)(V_(2)-V_(1))` But for adiabatic process `W=dU=((P_(2)V_(2)-P_(1)V_(1))/(gamma-1))` `PV=nRT " "rArr10xx10=nxx0.082xx273rArrn=4.47` moles `-P_(ext)(V_(2)-V_(1))=((P_(2)V_(2)-P_(1)V_(1))/(gamma-1))...
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