Question

10 gram of oxygen is introduced into an evacuated vessel of 5dm3 capacity maintained at 27°c calculate pressure in NM -2

Answer 1

Answer:  $156040.5Nm^{-2}$

Explanation:

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = ?

V= Volume of the gas = $5dm^3$ = 5 L     {tex](1dm^3=1L)[/tex]

T= Temperature of the gas = 27°C = (27+273) K = 300 K

R= Gas constant = 0.0821 atmL/K mol

n=  moles of gas= $\frac{\text {given mass}{\text {Molar mass}}=\frac{10g}{32g/mol}=0.3125$

Putting in the values:

$P=\frac{nRT}{V}=\frac{0.3125\times 0.0821\times 300}{5}=1.54atm$

$1 atm = 101325Nm^{-2}$

$1.54atm=\frac{ 101325}{1}\times 1.54=156040.5Nm^{-2}$

Thus pressure will be $156040.5Nm^{-2}$

Answer 2

Answer:

Answer:l hope it will help you

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