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

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)

Ask Question

Tag:MathPhyChemBioOther

Take Test

...