Question

What is the volume, in l, of 0.682 moles of o'what 2 ​ gas having a temperature of 68.2 ºc and pressure of 5.82 atm?

Answer 1

Using the Ideal Gas Law, the volume of 0.682 moles of oxygen gas at 68.2 °C and 5.82 atm pressure is calculated to be 3.241 liters.

Step-by-step explanation:

To calculate the volume of the oxygen gas (O2), we can use the Ideal Gas Law, which is PV = nRT, where P is the pressure in atmospheres (atm), V is the volume in liters (L), n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin (K).

First, convert the temperature from Celsius to Kelvin by adding 273.15. Thus, T = 68.2 °C + 273.15 = 341.35 K.

Now we can use the Ideal Gas Law:
(5.82 atm) × V = (0.682 moles) × (0.0821 L·atm/K·mol) × (341.35 K).

To find the volume, V, rearrange the equation: V = (nRT) / P and compute V using the given values:
V = (0.682 moles × 0.0821 L·atm/K·mol × 341.35 K) / 5.82 atm = 3.241 L.

Therefore, the volume of the 0.682 moles of oxygen gas at 68.2 °C and 5.82 atm pressure is 3.241 liters.

Answer 2

We will use Ideal gas equation. PV=nRT where P= pressure(in atm)=5.82atm V=Volume=? n= moles of gases=0.682 R = gas constant= 0.0821L atm / mol K T = temperature= 68.2 C= 68.2 + 273.15 = 341.35 K V= nRT/ P= 0.682 X 0.0821 X 341.35/ 5.82 V= 3.284 L.