At constant temperature the pressure of 22.4 dm3 volume of an ideal gas was increased from 105 kPa to 210 kPa, New volume could be -
a. 44.8 dm3
b. 11.2 dm3
c. 22.4 dm3
d. 5.6 dm3
1 Answers
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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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When an ideal gas is heated at constant pressure, the fraction of the heat energy supplied whicn increases the internal energy of the gas is :
Correct Answer - D `{:(dQ=dW+dU,,,,),(dQ=PdV+dU,,,,),(dQ=nRdT+dU,,,,):}` `dQ=(2dU)/(f)+dU" "[dU=(nfRdT)/(2)]` `(dU)/(dQ)=(1)/(((2)/(f)+1))` `(dU)/(dQ)=(5)/(7)`
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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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When two mole of an ideal gas `(C_(p.m) = (5)/(2)R)` heated from `300 K` to `600 K` at constant pressure. The change in entropy of gas `(DeltaS)` is:
Correct Answer - C `DeltaS = nC_(p.m) "In" (T_(2))/(T_(1)) = 2 xx (5)/(2)"R In" (600)/(300) = 5"R In 2`
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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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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 wil
Correct Answer - `525 K`
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The energy of alpha particles emitted by `.^(210)Po` is 5.3 MeV. The half life of this alpha emitter is 138 days. (a) What mass of `.^(210)Po` is need
Correct Answer - [88.4 mg, 0.16 watt]
2 Answers 2 views
Calculate the number of molecules of methane in 0.50 m^3 of the gas at a pressure of 2.0 × 10^2 kPa and a temperature of exactly 300 K.
Given : V = 0.5 m3, P = 2.0 × 10 kPa = 2.0 × 105 Pa T = 300 K, R = 8.314 J K-1 mol-1 To find : Number of molecules of methane gas Formula...
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हेप्टेन एवं ओक्टेन एक आदर्श विलयन बनाते है । 373k पर दोनों द्रव घटकों के वाष्प दाब क्रमश : 105.2 kPa तथा 46.5 kPa है । 26.0 g हेप्टेन एवं 35.0 g ओक्टे
Correct Answer - `73.58 kPa`
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