Calculate mass % of He in a mixture of `O_(2)` and He gas at 3 atm and `27^(@)C` havin density `(5)/(3)` gm/litre [Take : R=0.08 atm litre `mol^(-1)k^(-1)`]
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Assume liquefied petroleum gas (LPG) is a 50-50 (by mole) mixture of n-pentane and n-butnane.Calculate the calorific value (in `KJ//"mol"`) of gas ava
Correct Answer - 30 KJ/mol, increase 30 kJ/mol , increase First of all we find mole fractions in liquid mixture, let mixture is 100 mole So moles of pentane=50 moles of...
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138 gm of `N_2O_4(g)` is placed in 8.2L container at 300 K.The equilibrium vapour density of mixture was found to be 30.67.Then (R=0.082 L atm `"mol"^
Correct Answer - B,C,D `{:(,N_2O_4" "hArr,2NO_2),(t=0,a,0),(t,a(1-alpha),2aalpha):}` vapour density`=46/(1+alpha)=30.67` so `1+alpha=1.5=0.5=50%` Total pressure`=(1.5xx1.5xx0.082xx300)/8.2=6.75` atm So, `K_p=(4alpha^2)/(1-alpha^2)P=9` atm and for density of mixture `=138/8.2` gm/L=16.83 gm/L
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At a certain temperature for which RT=25 lit.atm. `"mol"^(-1)`., the density of a gas, in gas `lit^(-1)`, is d=2.00 P+0.020 `P^2`, where P is the pres
Correct Answer - B Lt `d/p=M/(RT)implies M=2RT=50` `Pto0`
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A spherical balloon of volume 5 litre is to be filled up with `H_2` at NTP from a cylinder of 6 litre volume containing the gas at 6 atm at `0^@C`.The
Correct Answer - 6 Volume of gas at NTP is , `P_1V_1=P_2V_2` `6xx 6 =1xxV_2` `V_2`=36 litres. Outcoming gas = 36-6 =30 litres No of balloon =`30/5=6`
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In a reaction, the concentration of a reatant (A) changes from `0.200 "mol litre"^(-1) to 0.150 "mol litre"^(-1)` in 10 minutes. What is the average r
`Delta[A]=[A]_("initial")` `=[0.150-0.200]=0.050 "mole litre"^(-1)` `Deltat=10` minutes Average rate of reaction `=(-Delta[A])/(Deltat)=(-[-0.050])/(10)` `=(0.050)/(10)=0.050 "mol litre"^(-1)"min"^(-1)`
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Define self -inductance and its `S.I.` unit. Derive an expression for self- industance of a long, air-cored solenoid of length , radius `r`, and havin
The self inductance of a coil is equal to the induced `emf` set up in the coil, when the current passing through it change at the unit rate The `SI`...
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The values of a and b for oxygen are `a = 1.360 atm litre^(2) mol^(-2)` and `b = 0.03183 litre//mol`. (i) Calculate the value of `(b-(a)/(RT)) at 0^(@
Correct Answer - `T ~~ 420 K`. (i) `b-(a)/(RT) = (0.03183 -(1.36)/(0.082xx273)) =- 0.029`. (ii) `:.` At `T_(B( (b-(a)/(RT)) = 0, i.e. b - (a)/(RT_(B)) = 0 rArr T_(B) = (a)/(Rb)...
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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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10 litre of `O_(2)` gas is reacted with 30 litre of CO gas at STP. Then volumes of each gas present at the end of reaction areL
Correct Answer - A `CO(g)+(1)/(2)O_(2)(g) to CO_(2)(g)` `{:(,t=0,30L,10L,0),(,"After reaction",(30-20)L,0,20L):}`
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For 0.2 mol `C_(3)H_(4)` and 0.5 mol `C_(2)H_(6)` gas mixture? Mass of mixture is 23 g It includes 1.6 mol C atom It includes `0.7 xx 6.022 xx 10^(23)
Correct Answer - C Mass of mixture `= 0.2 xx40+0.5 xx30=23 gm` Mole of c-atom `=0.2xx3+0.5xx2=1.6` mole Number of molecule `=(0.2 +0.5)N_(A)`
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