Question

The pressure exerted by $12\text{gm}$ of an ideal gas at $t^{\circ }\text{C}$ in a vessel of volume v litre is one atm. When the temperature is increased by ${10}^{\circ }\text{C}$ at the same volume, the pressure increases by 10%. Calculate the temperature t (in K) (Mol. Wt. of gas = 120)

Answer 1

Case- I $P=1\text{atm}w=12\text{gm}T=(t+273)\text{K}$
$V=v\text{litre}$
Case - II $T=(t+283)\text{K}P=1+\frac{10}{100}=1.1\text{atm}$

For case I
$PV=nRT$
$Numberofmoles{}^{\prime }{n}^{\prime }=\frac{Givenmass{}^{\prime }{w}^{\prime }}{Molecularmass{}^{\prime }{m}^{\prime }}$
So, $PV=\frac{w}{m}RT$
$V=\frac{12}{m}R(t+273)$

For case II
$1.1\text{V}=\frac{12}{m}\times R(t+283)$

For case (I) and (II)
$\frac{1.1}{1}=\frac{t+283}{t+273}$
$1.1t+300.3=t+283$
$t=-{173}^{\circ }\text{C}=(273+(-173\left)\right)\text{K}=100\text{K}$