The pressure of a gas is changed at constant volume from `200 Nm^(-2)` to `3000 Nm^(-2)`. If the initial temperature of the gas is `77^(@)C`, what will be its final temperature?
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One litre gas at `400K` and `300atm` pressure is compressed to a pressure of `600 atm` and `200K`. The compressibility factor is changed from `1.2 to
Correct Answer - `(1)/(3)litre` `Z_(i) = (P_(1)V_(1))/(RT_(1))` and `Z_(2) = (P_(2)V_(2))/(RT_(2))` `(Z_(1))/(Z_(2)) = (P_(1))/(P_(2)) xx (T_(2))/(T_(1)) xx (V_(1))/(V_(2)) rArr V_(2) = (1)/(3) litre`
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The specific heat at constant volume for a gas is 0.075 cal/g and at constant pressure it is 0.125 cal/g. Calculate, (i) the molecular weight of gas,
(i) `C_(P)-C_(V)=(R)/(M)` where, M=molecular of gas `0.125-0.075=(1.987)/(M)` `M=39.74~~40` (ii) `(C_(P))/(C_(V))=gamma` `therefore=(0.125)/(0.075)=1.66` `therefore` 1.66 value of `gamma` shows that the gas is monoatomic.
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`1` mole of an ideal diatomic gas undergoes a reversible polytropic process `(PV^(2)="constant")`. The gas expand from initial volume of `1` litre and
Correct Answer - B For polydropic process `(n=2)` `PV^(2)=C` `TV=C` `T_(1)V_(1)=T_(2)V_(2)` `300xx1=T_(2)xx3 rArrT_(2)=100 K` ` Delta U=nC_(v,m)Delta T=1xx(fR)/(2)(T_(2)-T_(1))=1xx(5xxR)/(2)xx(100-300)=(1xx5xxR)/(2)xx200= -500 R= -4.2 KJ`
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There is `1 mol` liquid (molar volume `100 ml`) in an adiabatic container initial, pressure being `1` bar Now the pressure is steeply increased to `10
Correct Answer - `Delta H=990, Delta E=19 J` `Delta U= q+W` for adiabatic process `q=0`, hence `Delta U=W` and `W= -p(Delta V)= -P(V_(2)-V_(1))` Now `Delta H= Delta U+Delta (PV)` Here `Delta...
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When one mole of an ideal gas is compressed to half of its initial volume and simultaneously heated to twice its initial temperature , the change in e
Correct Answer - D `DeltaS =nC_(v.m) "In" (T_(2))/(T_(1)) + nR "In" (V_(2))/(V_(1)) = C_(v.m) "In" 2+ R "In" ((1)/(2)) = (C_(v.m)-R) In 2`
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An ideal gas at initial temperature 300 K is compressed adiabatically (`gamma = 1.4`) to `(1/16)^(th)` of its initial volume. The gas is then expanded
Correct Answer - 1819K
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The value of `((100),(0))((200),(150))+((100),(1))((200),(151))+......+((100),(50))((200),(200))` equals (where `((n),(r ))="^(n)C_(r)`)
Correct Answer - A `(a)` Writing `((200),(r ))=((200),(200-r))`, we have `((100),(0))((200),(50))+((100),(1))((200),(49))+((100),(2))((200),(48))+......+((100),(50))((200),(0))` `=` Coefficient of `x^(50)` in the expansion of `(1+x)^(100)(x+1)^(200)` `=` Coefficient of `x^(50)` in the binomial expansion of `(1+x)^(300)` `=((300),(50))`
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Find Q1 , D5 and P45 for the following data: Monthly Rent 0-500 500-1000 1000-1500 1500-2000 2000-3000 3000-4000 4000-5000 5000-6000 6000-7000
Q1 = 1862.64, D5 = 2666.67, P45 = 2500
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At constant temperature the pressure of 22.4 dm^3 volume of an ideal gas was increased from 10^5 kPa to 210 kPa, New volume could be -
Option : b. 11.2 dm3
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A balloon is inflated with helium gas at room temperature of 25 °C and at 1 bar pressure when its initial volume is 2.27L and allowed to rise in air.
Given : P1 = Initial pressure = 1 bar V1 = Initial volume = 2.27 L P2 = Final pressure = 0.3 bar To find : V2 = Final volume Formula : P1V1 = P2V2 (at constant...
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