Question

2.0 mole of N2 gas is kept in a vessel of Volume 0.0 20 m cube and the temperature was maintained at 60 degree Celsius calculate the pressure of the gas (assumed that N2 gas is ideal)

Answer 1

n=2 mol

V=0.02 m³

T= 60° = 273 + 60 = 333 K

R= 8.314 JK-¹

We know that ,

PV=nRT

so, P= nRT÷V

P = 2 x 8.314 x 333 ÷ 0.02

P= 276,856.2 Nm-²

Answer 2

The pressure of the gas is 2.73 atm

Explanation:

To calculate the pressure of the gas, we use the equation given by ideal gas, which is:

$PV=nRT$

P = pressure = ?

V = volume = $0.020m^3=20.0L$     (Conversion  factor:  $1m^3=1000L$ )

n = number of moles = 2.0 mol

R = Gas constant = $0.0821\text{ L atm }mol^{-1}K^{-1}$

T = temperature = $60^oC=[60+273]K=333K$

Putting values in above equation, we get:

$p\times 20.0L=2.0\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 333\\\\P=\frac{2.0\times 0.0821\times 333}{20.0}=2.73atm$

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