Question

The amount of gas that occupies 60.82 l at 31.0 °c and 367 mm hg is ________ mol.

Answer 1

Answer:

1.17mole

Explanation:

Given parameters:

Volume of gas = 60.82L or 60.82dm³

Temperature of the gas = 31.0°C or 304K

Pressure on the gas = 367mmHg

To make calculation easier we convert the pressure from mmHg to atm.

           1 atm = 760mmHg

           367mmHg gives $\frac{367}{760}$atm = 0.48atm

Solution

Assuming ideality, we can find the number of moles of the gas used. The ideal gas law combines Boyle's law, Charles's Law and Avogadro's law. It is expressed below:

                   PV = nRT

P is the pressure

V is the volume

n is the number of moles

R is the gas constant given as 0.082atmdm³mol⁻¹K⁻¹

T is the temperature in kelvin

The unknown is n, which is the number of moles.

Making n the subject of the formula, we have ;

                    n = $\frac{PV}{RT}$

      n = $\frac{0.48 x 60.82}{0.082x304}$= 1.17mole

Answer 2

The amount of the gas is 1.17 moles.

What is the ideal gas law?

The ideal gas law is depicted by the relation; PV = nRT

P = pressure of the gas

V = volume of the gas

n = number of moles

R= gas constant

T = absolute temperature

Hence;

n = PV/RT

P = 367 mm hg or 0.48 atm

V = 60.82 l

T = 31.0 °c or 304 K

n = 0.48 atm * 60.82 l/0.082 atmL/Kmol *  304 K

n = 29.19/24.93

n = 1.17 moles

Learn more about ideal gas equation: https://brainly.com/question/1190311