Middle School

What is the pressure of 0.540 mol of an ideal gas at 35.5 L and 223 K?

Answer :

Answer:

  • 0.0443 atm

Explanation:

1) Data:

a) n = 0.540 mol

b) V = 35.5 liter

c) T = 223 K

2) Formula:

  • Ideal gas equation: PV = nRT

Where,

  • P = pressure
  • V = volume
  • n = number of moles
  • T = absolute temperature (kelvin scale)
  • R = universal gas constant = 0.08203 atm-liter/K-mol

3) Calculations:

  • Solve for P

PV = nRT ⇒ P = nRT/V

  • Substitute and compute:

P = 0.540 mol × 0.08206 atm-liter/K-mol × 223 K / 35.5 liter

P = 0.0443 atm (you must use 3 significant figures).

Using the ideal gas law PV = nRT, we calculated the pressure of 0.540 mol of an ideal gas at 35.5 L and 223 K. By substituting the given values into the formula, we found the pressure to be 28.2 kPa.

Use PV = nRT and R = 8.314 L kPa / mol K to solve.

We will use the ideal gas law formula: PV = nRT.

P is the pressure in kPa

V is the volume, which is 35.5 L

n is the number of moles, which is 0.540 mol

R is the ideal gas constant, which is 8.314 L kPa / mol K

T is the temperature, which is 223 K

Rearrange the formula to solve for P:

P = (nRT) / V

Substitute the given values into the equation:

P = (0.540 mol × 8.314 L kPa / mol K × 223 K) / 35.5 L

Calculate the value:

P = (1002.3362) / 35.5

P = 28.2 kPa

Therefore, the pressure of the gas is 28.2 kPa.