# Find the change in the Kinetic energy of an Ideal Gas

• Helly123
In summary, the answer to question 1 is that when the volume doubled at constant temperature, the kinetic energy per molecule became greater by 0.4. When the volume was doubled without any transfer of heat, the kinetic energy per molecule became greater by 1.4.
The first question is straightforward. Since the kinetic energy/molecule is (3/2)kT, the kinetic energy/mol is NA (3/2)kT =(3/2)RT. Further, this means that you are considering a monatomic gas which we know from kinetic theory has c(V) = 3R/2 and c(p) = 5R/2. The second question is for a V=con process, Q=?, since dV=0, Wk=0, and ΔU = Q, but ΔU=nc(V)ΔT=Q. The third question is for a p=con process, Wk=?, ΔU=Q-Wk, Wk=Q-ΔU=nc(p)ΔT-nc(V)ΔT=[c(p)-c(V)]nΔT=nRΔT=pΔV. The fourth question is for ΔT=0, Δε(kinetic)=?, since ε(kinetic)∝T, Δε(k)=0, The fifth question is for Q=0 and V(final)=2V(initial), an adiabatic expansion, ΔT=? The confusing part is the listed pressure ratio of p(final)=0.7p(initial). If it is truly an adiabatic expansion the relation p(f)/p(i)=[V(i)/V(f)]^γ=(1/2)^1.67=0.314. Maybe this is what was meant, p(f)=0.314p(i), just poorly worded, in that case T(f)/T(i)=[V(i)/V(f)]^(γ-1)=(1/2)^0.67=0.629=ε(k,f)/ε(k,i).

Helly123
Dr Dr news said:
The first question is straightforward. Since the kinetic energy/molecule is (3/2)kT, the kinetic energy/mol is NA (3/2)kT =(3/2)RT. Further, this means that you are considering a monatomic gas which we know from kinetic theory has c(V) = 3R/2 and c(p) = 5R/2. The second question is for a V=con process, Q=?, since dV=0, Wk=0, and ΔU = Q, but ΔU=nc(V)ΔT=Q. The third question is for a p=con process, Wk=?, ΔU=Q-Wk, Wk=Q-ΔU=nc(p)ΔT-nc(V)ΔT=[c(p)-c(V)]nΔT=nRΔT=pΔV. The fourth question is for ΔT=0, Δε(kinetic)=?, since ε(kinetic)∝T, Δε(k)=0, The fifth question is for Q=0 and V(final)=2V(initial), an adiabatic expansion, ΔT=? The confusing part is the listed pressure ratio of p(final)=0.7p(initial). If it is truly an adiabatic expansion the relation p(f)/p(i)=[V(i)/V(f)]^γ=(1/2)^1.67=0.314. Maybe this is what was meant, p(f)=0.314p(i), just poorly worded, in that case T(f)/T(i)=[V(i)/V(f)]^(γ-1)=(1/2)^0.67=0.629=ε(k,f)/ε(k,i).
Wow. Thanks a lot for reviewing all questions

Dr Dr news said:
The first question is straightforward. Since the kinetic energy/molecule is (3/2)kT, the kinetic energy/mol is NA (3/2)kT =(3/2)RT. Further, this means that you are considering a monatomic gas which we know from kinetic theory has c(V) = 3R/2 and c(p) = 5R/2. The second question is for a V=con process, Q=?, since dV=0, Wk=0, and ΔU = Q, but ΔU=nc(V)ΔT=Q. The third question is for a p=con process, Wk=?, ΔU=Q-Wk, Wk=Q-ΔU=nc(p)ΔT-nc(V)ΔT=[c(p)-c(V)]nΔT=nRΔT=pΔV. The fourth question is for ΔT=0, Δε(kinetic)=?, since ε(kinetic)∝T, Δε(k)=0, The fifth question is for Q=0 and V(final)=2V(initial), an adiabatic expansion, ΔT=? The confusing part is the listed pressure ratio of p(final)=0.7p(initial). If it is truly an adiabatic expansion the relation p(f)/p(i)=[V(i)/V(f)]^γ=(1/2)^1.67=0.314. Maybe this is what was meant, p(f)=0.314p(i), just poorly worded, in that case T(f)/T(i)=[V(i)/V(f)]^(γ-1)=(1/2)^0.67=0.629=ε(k,f)/ε(k,i).
So, p f = 0.3 p i
Not that pf 0.3 lower than pi?

That is what it looks like to me.

Dr Dr news said:
That is what it looks like to me.
Ok. Thanks sir

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