Question

Given is the graph between $\frac{PV}{T}$ and P for 1 g of oxygen gas at two different temperatures T1 and T2, as shown in figure. Given, density of oxygen = 1.427 kg m–3. The value of PV/T at the point A and the relation between T1 and T2 are respectively.

• A. 0.259 J K–1 and T1 < T2
• B. 8.314 g J mol–1 K–1 and T1 > T2
• C. 0.259 J K–1 and T1 > T2
• D. 4.28 g J K–1 and T1 < T2

Answer 1

Answer: (C)

0.259 J K–1 and T1 > T2

Answer 2

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

Where m = mass of the gas

And $\frac{m}{M} = \mu =$number of moles.

$\frac{PV}{T} = \mu R =$a constant for all values of P.

That is why, ideally it is a straight line.

$\therefore\frac{PV}{T} = \frac{1g}{32gmol^{- 1}}8.31Jmol^{- 1}K^{- 1} = 0.259JK^{- 1}$

Also $T_{1} > T_{2}$