JEE Main · 2023 · 13 Apr 2023 · Shift 1 · States of Matter

A certain quantity of real gas occupies a volume of

0.15 dm^{3}

at 100 atm and 500 K when its compressibility factor is 1.07. Its volume at 300 atm and 300 K (When its compressibility factor is 1.4) is

____

\times 1 0^{- 4} dm^{3}

(Nearest integer)

Answer: 392

Solution

For a real gas, the compressibility factor

Z = \frac{P V}{n R T}

, which gives

n = \frac{P V}{Z R T}

. Since the number of moles remains constant:

\frac{P _{1} V _{1}}{Z _{1} R T _{1}} = \frac{P _{2} V _{2}}{Z _{2} R T _{2}}

Substituting the values:

P_{1} = 100

atm,

V_{1} = 0.15

dm³,

Z_{1} = 1.07

T_{1} = 500

P_{2} = 300

atm,

Z_{2} = 1.4

T_{2} = 300

V_{2} = \frac{P _{1} V _{1} Z _{2} T _{2}}{P _{2} Z _{1} T _{1}} = \frac{100 \times 0.15 \times 1.4 \times 300}{300 \times 1.07 \times 500} = \frac{1.4 \times 0.15 \times 300}{1.07 \times 500} = 0.0392 dm^{3} = 392 \times 1 0^{- 4} dm^{3}

JEE Main · 2023 · 13 Apr 2023 · Shift 1 · Integer answer