Question

The pressure exerted by $12g$ of an ideal gas at temperature $1^{o}C$ in a vessel of volume $V$ litre is one atm. When the temperature is increased by $10$ degrees at the same volume, the pressure increases by $10$%. Calculate the temperature $t$ and volume $V$. (Molecular weight of the gas$=120$)

Answer 1

Case I : Given

P = 1 atm

w = 12 g

T = (t+273)K

V = V litre

Case ii :

T = (t+283)K

P = 1 + 1010010100 = 1.1 atm

w = 12 g

V = V l

Using gas equation

Case I : 1×V=12m×R(t+273)1×V=12m×R(t+273) ----(i)

Case II : 1.1×V=12m×R(t+283)1.1×V=12m×R(t+283) ----(ii)

By equation (i) and (ii)

1.11=t+283t+2731.11=t+283t+273

∴1.1t+300.3=t+283∴1.1t+300.3=t+283

0.1t = -17.3

t=−173∘Ct=−173∘C = 100K

Also from case I (Since m = 120)

1×V=12120×0.082×1001×V=12120×0.082×100

V = 0.82 l