Question

1.0 moles of pure dinitrogen gas SATP conditions was put in to a vessel
of volume 24.8 m cube, maintained at the temperature of 596.3 K. What is the
pressure of the gas in the vessel
1) 2 x 10-3 bar 2) 4 * 10-3 bar 3) 5 x 10-3bar 4) 2 x 10-1 bar​

Answer 1

Answer : The correct option is, (1) $2\times 10^{-3}bar$

Explanation :

Using ideal gas equation:

$PV=nRT$

where,

P = pressure of gas = ?

V = volume of gas = $24.8m^3=24800L$

conversion used : $(1m^3=1000L)$

T = temperature of gas = 596.3 K

n = number of moles of gas = 1.0 mole

R = gas constant = 0.08314 L.bar/mole.K

Now put all the given values in the ideal gas equation, we get pressure of the gas.

$P\times (24800L)=1.0mole\times (0.08314L.bar/mole.K)\times (596.3K)$

$P=1.99\times 10^{-3}bar\approx 2\times 10^{-3}bar$

Therefore, the pressure of the gas in the vessel is, $2\times 10^{-3}bar$

Answer 2

Answer:

the correct answer is option 1