Question

Calculate the volume occupied by 4.045 x 10^23 molecules of oxygen at 27°C and having a pressure of
0.935 bar​

Answer 1

Answer:

178.6 is your answer

Explanation:

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Answer 2

Answer:

The quantity occupied through 4.045 x 10^23 molecules of oxygen at 27°C and 0.935 bar strain is 0.0152 cubic meters

Explanation:

From the above question,

                      PV = nRT

where,

P is the pressure,

V is the volume,

n is the quantity of moles of gas,

R is the prevalent gasoline constant and

T is the temperature.

1 mole of gasoline carries 6.022 x $10^{23$ molecules

                n = (4.045 x $10^{23$ molecules) / (6.022 x $10^{23$ molecules/mol)

                n =  0.673 mol

The temperature is given as 27°C, which can be transformed to Kelvin by way of including 273.15

                T = 27°C + 273.15

                T = 300.15 K

                P = 0.935 bar x $10^5$ Pa/bar

                P = 93500 Pa

Now we can replacement these values in the perfect gasoline regulation to locate the volume:

                V = (nRT) / P

                V = (0.673 mol x 8.314 J/(mol·K) x 300.15 K) / 93500 Pa

                V = 0.0152 $m^3$

Therefore, the quantity occupied through 4.045 x 10^23 molecules of oxygen at 27°C and 0.935 bar strain is 0.0152 cubic meters (or 15.2 liters).

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