Question

Given is the graph between $\frac{PV}{T}$ and $P$ for $1g$ of oxygen gas at two different temperatures $T_1$ and $T_2$, as shown in figure. Given, density of oxygen $=1.427kg{m}^{-3}$.The value of $PV/T$ at the point $A$ and the relation between $T_1$ and $T_2$ are respectively.

• A. $0.259$J $K^{-1}$ and $T_1<T_2$
• B. $8.314$g J $mo{l}^{-1}$ $K^{-1}$ and $T_1>T_2$
• C. $0.259$J $K^{-1}$ and $T_1>T_2$
• D. $4.28$g J $K^{-1}$ and $T_1<T_2$.

Answer 1

The correct option is C $0.259$J $K^{-1}$ and $T_1>T_2$

For an ideal gas:

$PV=\mu RT=\frac{m}{M}RT$,

Here $m$ is the mass of the gas and $\frac{m}{M}i.e.\mu$ is the number of moles of gas.

The value of the equation remains constant for all values of $P$.

$\frac{PV}{T}=\mu R=a$

That is why ideally it is a straight line.
$\therefore \frac{PV}{T}=\frac{1g}{32gmo{l}^{-1}}\times 8.31J$

$mo{l}^{-1}{K}^{-1}=0.259$J $K^{-1}$

Since real gases behave ideally at low pressure and high temperature.

So,$T_1>T_2$