Q.
Pressure of 1 gm an ideal gas A at 27°C is found to be 2 bar. When 2 g of another ideal gas B is introduced in the same flask at same temperature, the pressure becomes 3 bar. Find a relationship between their molecular masses.
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a
MB=3MA
b
MB=4MA
c
MB=2MA
d
MB=MA
answer is B.
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Detailed Solution
pV = nRTFor A gas, pAV=nART ...(i)Similarly for B gas, pBV=nBRT -----(ii)Number of moles of A gas ;nA=1MAMA= molar mass of gasA Number of moles of B gas; nB=2MBMB= molar mass of gas BPressure of gas A, pA=2 bar Total pressure, ptotal =pA+pB=3 bar Pressure of gas B, pB=ptotal -pA=3-2=1barV , R and T are same for both the gases.Hence, from Eqs.(i) and (ii), we get pApB=nAnB=1×MBMA×2MBMA=2pApB=2×21MB=4MAAlternate method(b) (i) ∵pT=pA+pBGiven, pT=3 bar and pA=2 bar ∴ pB=3-2=1bar (ii) ∵pV=nRT=wMRT∴ For (A)2=1MA---------(I)∵pA=2 bar and V,T are constant. wA=1 g,MA= molar mass of ASimilarly for (B)1=2MB --------(ii)∵pB=1bar and V,T are constant wB=2 g,MB= molar mass of MB ) Dividing (i) by (ii)21=1MA2MB=MB2MA, i.e. MB=4MA
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