Question

At NTP volume of 1kg air is 0.773 m3
. Calculate
the gas constant of air. If pressure and temperature
of air are 10 bar and 400°C respectively. So find
mass of air if volume of air is 3 m3.

Answer 1

Answer:

First we derive an expression for density of a gas.

Density =

volume(V)

mass(w)

but we know that PV = nRT

now n =

M

w

where M is molecular weight

So, PV =

M

w

RT

this gives

V

w

=

RT

PM

= density = d

Clearly at a constant temperature,

P

1

d

1

=

P

2

d

2

so since pressure is tripled, density also triples

new density = old density x 3

=1.293×3

=3.87gm/ltr

Answer 2

The mass of air is 3.88 kg.

Given:
Mass of air $=1kg$

Volume of air $=0.773mm^{3}$

Pressure $=10bar$

Temperature $=400^{o}C$

To find:

Gas constant of air.

Solution:

Step 1

We have been given a system where,

$m=1kg=1000g$

$V=0.773mm^{3}$

$P=10bar=10^{6}N/m^{2}$

$T=400^{o} C=673K$  

We know, the ideal gas equation for gases states that,

$PV=nRT$

Where, $n$ is number of moles given by the ratio of the mass $(m)$ of the substance and its molecular weight $(M)$.

$n=\frac{m}{M}$

We know, the molecular weight of air is $M=28.96g/mol$.

Substituting the known values in the equation, we get

$10^{6}(0.773) =$ $\frac{1000}{28.96}$  × $R$ × $673$

$R=\frac{773*28.96}{673}$

$R=\frac{22386.08}{673}$

$R=33.2$ $Pa.m^{3}/K /mol$

Step 2

Now,

We have, $V=3m^{3}$

Substituting this value in the equation, we get

$10^{6}(3)=\frac{m}{28.96}(33.2) (673.15)$

$10^{6}(3)=m$ × $771.8$

$m=3.88kg$

Final answer:

Hence, the mass of the air is 3.88 kg.