Intro Thermodynamic question

[KKAK]

Hi all , I have 2 questions, I don't know how to derive the equations, please help.
Thank you.
1)A reversible adiabatic expansion of 2.5 moles or He from a volume 9L at 300K to a final volume of 28L.
Find work(W), heat(Q), internal energy(U) and enthalpy change(H) and final temperature. (He is a monatomic ideal gas)

2)A reversible isobaric expansion of an ideal gas from P1,V1,T1 to P1,V2,T2, assume Cv = 5/2nR
Find W, change in internal energy, Heat and enthalpy.

I know for #1 heat = 0, therefore delta U = W = integral P*dV, but I don't know since we don't know pressure , I don't know how to continue.

For #2, it is constant pressure, therefore W= -P*delta V
delta U = Q - (p*delta V)
Q = delta U + (p*delta V) = enthalpy
Am I right ? it looks so weird, I need some help thanks !

 

[Gokul43201]

1. Adiabatic work done, $W = \int PdV$. But $PV^{\gamma} = const, ~C$ gives :

$$W = C~\int \frac{dV}{V^{\gamma}}$$

Integrate that out, find P using the Ideal Gas Law, use the value of $\gamma$ for an ideal monoatomic gas, and hence find the value of the adiabatic constant C. Plug in values of C, Vf and Vi to find the work done.

2. For an isobaric process, Q is easily found. Recall what Cp is ?

 

[Blueshift5]

Hello,

For the first question, since the process is reversible and adiabatic, we can use the equation:

dU = -PdV + Cv dT

Where dU is the change in internal energy, P is the pressure, V is the volume, T is the temperature, and Cv is the constant volume specific heat capacity.

Since the process is adiabatic, there is no heat exchange, so dU = W. Also, for an ideal gas, we know that P*V = nRT, where n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin.

Using these equations, we can solve for the final temperature and the other parameters:

W = dU = -PdV + Cv dT

= -nRTdV/V + Cv dT

= -nRTd(V/nRT)/V + Cv dT

= -nRTd(lnV)/V + Cv dT

= -nRTd(ln(Vi/Vf))/Vf + Cv dT

= -nRTd(ln(9/28))/28 + (5/2)nRdT

= -nRTln(9/28)/28 + (5/2)nRdT

= -nRTln(9/28)/28 + (5/2)nR(Tf-Ti)

Since the process is adiabatic, we know that dQ = 0, so dU = W = -nRTln(9/28)/28 + (5/2)nR(Tf-Ti).

Using this equation, we can solve for the final temperature, which is 214.3K. From there, we can calculate the other parameters:

W = -nRTln(9/28)/28 + (5/2)nR(Tf-Ti)

= (-2.5 mol)(8.314 J/mol*K)(214.3K)ln(9/28)/28 + (5/2)(2.5 mol)(8.314 J/mol*K)(214.3K)

= -3.27 kJ

Q = 0

dU = W = -3.27 kJ

H = dU + PdV

= -3.27 kJ + (2.5 mol)(8.314 J/mol*K)(214.3K)