If I have 1.9 moles of gas held at a pressure of 4.95 atm and in a container with a volume of 5000.0 mL, what is the temperature of the gas?

158.66 K

Explanation:

Given that:

Moles of the gas = 1.9 moles

Pressure = 4.95 atm

Volume = 5000.0 mL

Also, 1 mL = 10⁻³ L

So, volume = 5 L

Using ideal gas equation as:

PV=nRT

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L. atm/K. mol

Applying the equation as:

4.95 atm * 5 L=1.9 moles * 0.0821 L. atm/K. mol * T

⇒T = 158.66 K