A gaseous chemical has a concentration of 41.6 \mu {mol} {/} {m}^{3} in air at 1 atm pressure and temperature 293 {K}. The universal gas constant {R} is 82.05 \times 10^{-6}( {m}^{3} {atm}) \/( {mol} {K}). Assuming that ideal gas law is valid, the concentration of the gaseous chemical (in ppm, round off to one decimal place), is ______________.

GATE Civil Engineering (CE) 2020 Shift 1 Solved Paper

Question : 41 of 65

Marks: +1, -0

A gaseous chemical has a concentration of

41.6 µ {mol} {/} {m}^{3}

in air at 1 atm pressure and temperature

293 {K}

. The universal gas constant

{R}

82.05 × 10^{-6}( {m}^{3} {atm}) \/( {mol} {K})

.
Assuming that ideal gas law is valid, the concentration of the gaseous chemical (in ppm, round off to one decimal place), is ______________.

Solution:

We know that,

{PV}= {Nrt}

Where,

{P}=

Pressure

{V}=

Volume

n=

number of moles

{R}=

Universal gas constant

{T}=

Temperature
So,

V=\;{n R T}/{P}

=\;{41.6 × 10^{-6} × 32.05 × 10^{-6}}/{1} × 293

⇒ {V}=10^{-6} {m}^{3}

⇒ {V}=10^{-6} {m}^{3}

1 {ppm}=\;{1 \;\text " part of gas "\;}/{10^{6} \;\text " parts of air "\;}=\;{1 {m}^{3} \;\text " of gas "\;}/{10^{6} {m}^{3} \;\text " of air "\;}

⇒ 1 {ppm}=\;{41.6 × 10^{6}-4 {\moles}}/{10^{6} {cm}^{3}}

Since,

41.6-4

mole of gas volume of

10^{-6} {m}^{3}

∴ 41.6-4

moles

{/} {m}^{3}=1 {ppm}

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