Question

pressure exerted by 16g of oxygen gas in 0.01 m^3 vessel at 47 degree celcius is?​

Answer 1

The pressure exerted by 16g of oxygen gas in a 0.01 m^3 vessel at 47 degrees Celcius is 6502 atm.

Given:

The pressure exerted by 16g of oxygen gas in n 0.01 m^3 vessel at 47 degrees Celcius.

To Find:

The pressure was exerted by 16g of oxygen gas in a 0.01 m^3 vessel at 47 degrees Celcius.

Solution:

To find the pressure exerted by 16g of oxygen gas in a 0.01m^3 vessel at 47 degrees Celcius we will follow the following steps:

As we know,

PV = nRT

Here, p is the pressure in atm and v is the volume in m³.

T is the temperature in kelvin = 273 + 47 = 720 K and R is the gas constant = 0.0821 atm L.

1 mole of oxygen = 16 grams.

the is the number of moles =

$\frac{16}{16} = 1$

Now,

Rearranging and putting values in the above formula we get,

$p = \frac{nRT}{v} = \frac{1 \ time 0.0821 \times 792}{0.01} = 6502atm$

Henceforth, the pressure exerted by 16g of oxygen gas in a 0.01 m^3 vessel at 47 degrees Celcius is 6502 atm.