What is the pressure exerted by 1.2 mol of a gas with a temperature of 20°C and a volume of 9.5 L?

Question

Ellie Andrews · Accepted Answer

#P~=3*atm#

All we do is apply the Ideal Gas equation.

#P=(nRT)/V# and use an appropriate gas constant, #R=0.0821*L*atm*K^-1*mol^-1#...and of course we use absolute temperature....

#P=(1.2*molxx0.0821*(L*atm)/(K*mol)xx293*K)/(9.5*L)=??*atm#

Cooper Adams · Answer

undefined To find the pressure exerted by the gas, you can use the ideal gas law formula: $ PV = nRT $, where $ P $ is the pressure, $ V $ is the volume, $ n $ is the number of moles of the gas, $ R $ is the ideal gas constant, and $ T $ is the temperature in Kelvin. First, convert the temperature from Celsius to Kelvin by adding 273.15:

$ T = 20°C + 273.15 = 293.15 \, K $

Then, plug in the given values:

$ n = 1.2 \, mol $

$ V = 9.5 \, L $

$ T = 293.15 \, K $

The ideal gas constant, $ R $, is 0.0821 L·atm/(mol·K).

Now, solve for $ P $:

$ P = \frac{nRT}{V} $

$ P = \frac{(1.2 \, mol)(0.0821 \, L·atm/(mol·K))(293.15 \, K)}{9.5 \, L} $

$ P ≈ 3.86 \, atm $

So, the pressure exerted by 1.2 mol of the gas at a temperature of 20°C and a volume of 9.5 L is approximately 3.86 atm.