Question

A pressure cooker contains 0.355 mole of air, which occupies a volume of 5.68 liters. The temperature inside the pressure cooker is 394 K. What is the absolute pressure of the air in the pressure cooker? The pressure of the air is pascals.

Answer 1

The problem can be solved using Ideal gas Equation

PV = nRT

P = Pressure in atm (can be converted to pascals)

V = Volume in Liters

N stands number of moles

R is the gas constant 0.08206 L atm
`K^(-1) mol^(-1)`

T is the temperature in kelvin

`$P=(n R T)/(V)=\frac{0.355 \mathrm{mol} * 0.08206 \mathrm{LatmK}^(-1) \mathrm{mol}^(-1) * 394 \mathrm{K}}{5.68 \mathrm{L}} $$P=2.021 \mathrm{atm} $`

We know 1 atm = 101.325 kilo pascals = 101325 pascals

`$2.021 {atm} * \frac{101325 \text { ascals }}{1 \text { atm }}=204750 \text { Pascals } $$=2.05 * 10^(5) \text { Pascals (Answer) } $`

Answer 2

The pressure of air is 204848.75 pascal.

Step-by-step explanation:

Given data:

Moles of air = 0.355 mole

Volume of air = 5.68 L

Temperature = 394 K

Solution:

We will solve this problem by using ideal gas equation i.e.

PV = nRT

R is the ideal gas constant = 0.0821 L. atm / mol.K

Now we will put the values in equation,

PV = nRT

P = nRT / V

P = (0.335 mol × 0.0821 L. atm / mol.K × 394 K ) / 5.68 L

P = 11.483 L . atm / 5.68 L

P = 2.0217 atm

Now we will convert the atm into pascal.

1 atm = 101325 pa

2.0217 × 101325 = 204848.75 pa