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The pressure exerted by 12g of an ideal gas at temperature $t^{\large\circ}C$ in a vessel of volume V litre is one atm. when the temperature is increased by 10 degree at the same volume , the pressure increases by $10\%$. Calculate the temperature t and volume V (molecular weight of the gas = 120 )

$(a)\;100K ; 0.82\;l\qquad(b)\;110K;1.32\;l\qquad(c)\;96K;1\;l\qquad(d)\;105K;2.3\;l$

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Case I : Given
P = 1 atm
w = 12 g
T = (t+273)K
V = V litre
Case ii :
T = (t+283)K
P = 1 + $\large\frac{10}{100}$ = 1.1 atm
w = 12 g
V = V l
Using gas equation
Case I : $1\times V = \large\frac{12}{m}\times R(t + 273)$ ----(i)
Case II : $1.1\times V = \large\frac{12}{m}\times R(t + 283)$ ----(ii)
By equation (i) and (ii)
$\large\frac{1.1}{1} = \large\frac{t + 283}{t + 273}$
$\therefore 1.1t + 300.3 = t + 283$
0.1t = -17.3
$t = -173^{\large\circ}C$ = 100K
Also from case I (Since m = 120)
$1\times V = \large\frac{12}{120}\times0.082\times 100$
V = 0.82 l
Hence answer is (A)
answered Apr 4, 2014 by sharmaaparna1
 

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