one mole of ideal gas is slowly heated at a constant pressure of 2.0 atm from 200 to 300k.Calculate
q,w,£GE,and£GH,allinjoules.AssumeCp=5 2 R
p =1 atm
n =2
Cp =(5/2)R
£GT =300-200 = 100 k
so Cv =Cp -R = (3/2)R
q =nCp£GT= 2(5/2)R*100 = 500R =500*8.314 =4157 j
£GE =nCv£GT= 2(3/2)R*100 = 300R =2494.2 j
£GH =nCp£GT =q =4157 j
it' that right??
Your calculations for q, £GE, and £GH are correct, but there is a mistake in calculating the value of £GE. Let's go through the calculations again.
Given:
Pressure (p) = 2 atm
Number of moles (n) = 1 (since one mole of ideal gas is given)
Initial temperature (T1) = 200 K
Final temperature (T2) = 300 K
Cp = (5/2)R
To calculate q (heat absorbed at constant pressure), we use the formula:
q = n * Cp * ΔT
Substituting the values:
q = 1 * (5/2)R * (300 - 200)
q = 1 * (5/2) * 8.314 J/mol K * 100 K
q = 2078.5 J
So, the correct value of q is 2078.5 J.
To calculate £GE (change in Gibbs free energy), we use the formula:
£GE = n * Cv * ΔT
where Cv is the molar heat capacity at constant volume and is given by Cp - R.
Cv = (5/2)R - R
Cv = (3/2)R
Substituting the values:
£GE = 1 * (3/2)R * (300 - 200)
£GE = 1 * (3/2) * 8.314 J/mol K * 100 K
£GE = 1247.1 J
So, the correct value of £GE is 1247.1 J.
Lastly, £GH is equal to q since the process is at constant pressure, so £GH = q. Therefore, the correct value of £GH is 2078.5 J.
The corrected values are:
q = 2078.5 J
£GE = 1247.1 J
£GH = 2078.5 J