There are `4 xx 10^(24)` gas molecules in a vessle at `50 K` temperature. The pressure of the gas in the vessel is `0.03 atm`. Calculate the volume of the vessel.
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If a gas at a pressure of `10 atm` at `300 K` expands against a constant external pressure of `2 atm` from a vol. `10` litres to `20` litres find work
Process is irreversible `w= (-)underset(10)overset(20)int2dv= -2[20-10]=20 L. atm` `1` litre at `=101.3 J`
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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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One mole of an ideal monoatomic gas at temperature `T` and volume `1L` expands to `2L` against a constant external pressure of one atm under adiabatic
Correct Answer - B `Delta U=W` `nCV(T_(2)-T)= -Pxx(V_(2)-V_(1))` `(3)/(2)R(T_(2)-T)= -1 " " :. T_(2)=T-(2)/(3xx0.0821)`
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Following reaction occurs at `25^(@)` : `2NO(g, 1xx10^(-5)atm) + CI_(2)(g, 1 xx 10^(-2)atm)hArr 2NOCI(g,1 xx 10^(-2)atm)` Calculate `DeltaG^(@) [R = 8
Correct Answer - `-43.92 KJ` `2NO (9,1xx 10^(-5) atm) + CI_(2)(9, 1 xx 10^(-2)) hArr 2NOCI(9, 1 xx 10^(-2))` `DeltaG^(@) =- 2.363 "RT" ("log" ((1 xx 10^(-2))^(2))/((1 xx 10^(-5))^(2) (1 xx...
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A sample of certain mas s of an ideal polyatomic gas is expanded against constant pressure of 1 atm adiabatically from volume 2L , pressure 6 atm and
Correct Answer - 6 `W=-P_("ext")(V_(f)-V_(i))` =- (1 atm) (8-2)L =- 6L atm as q=0 so `DeltaE=w " " implies " " 3(8P_(f) -12)=-6` `8P_(f) =12 -(6)/(3)=10 implies P_(f)=(5)/(4) atm` so `(T_(f))/(T_(i))=((5)/(4)xx3)/(6xx2)=(10)/(12)`...
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A gas is filled in a vessel at a certain temperature and at a pressure of `80 cm` of Hg. At the same temperature, more gas is filled in the vessel so
Correct Answer - `128 cm` of Hg
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There are `6 xx 10^(21)` hydrogen molecules in a vessel of volume `200 cm^(3)`. Its tempera-ture is `27^(@)C` and the pressure is `10^(5) Nm^(-2)`. If
Correct Answer - (i) It remains unchanged `(3 xx 10^(19) "per cm"^(3))` , (ii) `15 : 16` , (iii) `15 : 16`
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Nitrogen gas is filled in a container of volume 2.32 L at 32°C and 4.7 atm pressure. Calculate the number of moles of the gas.
Given : V = 2.32 L, P = 4.7 atm, T = 32°C = 32 + 273.15 K = 305.15 K R = 0.0821 L atm K-1 mol-1 To find : n = number of moles of...
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