Chap. 15, #2. (I) A gas is enclosed in a cylinder fitted with a light frictionless piston and maintained at atmospheric pressure. When 1400 kcal of heat is added to the gas, the volume is observed to increase slowly from $12.0 m^3$ to $18.2 m^3$ . Calculate (a) the work done by the gas and (b) the change in internal energy of the gas.
The transformation is at constant pressure. $P = 1 atm = 10^5 N/m^2$
$Work = P*(V_f-V_i) = 10^5*(18.2-12.0) =6.2*10^5 Joule$
$Q = U + W$, Q is the heat, U is the change in the internal energy, W is the work
$1 cal =4.18 J$
$U = Q – W =1400*10^3*4.18 -6.2*10^5 =5.232*10^6 Joule$
Chap. 15, #4. (I) Sketch a PV diagram of the following process: 2.0 L of ideal gas at atmospheric pressure are cooled at constant pressure to a volume of 1.0 L, and then expanded isothermally back to 2.0 L, whereupon the pressure is increased at constant volume until the original pressure is reached.
Chap. 15, #8. (II) An ideal gas expands at a constant total pressure of 3.0 atm from 400 mL to 660 mL. Heat then flows out of the gas at constant volume, and the pressure and temperature are allowed to drop until the temperature reaches its original value. Calculate (a) the total work done by the gas in the process, and (b) the total heat flow into the gas.
The work done on the constant volume transformation is zero.
The work done on the constant pressure transformation is
$W =P*(V2-V1) = 3*10^5*(0.66-0.4)*10^-3 = 78 Joule$
Total heat absorbed in the constant pressure transformation $Q_p = \nu*C_p*\Delta(T)$ ,
delta T is the variation of temperature, niu the numberof gas moles, Cp the specific heat at constant pressure
Total heat released in the constant volume transformation is
$Q_v = -\nu*Cv*\Delta(T)$ , $Delta(T)$ is the same
$Q = Q_p-Q_v = \nu*\Delta(T)*(C_p-C_v) = \nu*R*\Delta(T) = p*V_2- P*V_1 = P*(V_2-V_1) =$ $=W =78 Joule$
because $C_p-C_v = R$ and $P*V = \nu*R*T$
Chap. 15, #10. (II) Consider the following two-step process. Heat is allowed to flow out of an ideal gas at constant volume so that its pressure drops from 2.2 atm to 1.4 atm. Then the gas expands at constant pressure, from a volume of 6.8 L to 9.3 L, where the temperature reaches its original value. See Fig. 15–22. Calculate (a) the total work done by the gas in the process, (b) the change in internal energy of the gas in the process, and (c) the total heat flow into or out of the gas.
Total work = work on the constant pressure transformation =
$= P*(V_2-V_1)= 1.4*10^5*(9.3-6.8)*10^-3 = 350 J$
$Total heat = -Q_v +Q_p = \nu*C_p*\Delta(T) – \nu*C_v*\Delta(T) = \nu*\Delta(T)*(C_p-C_v) =$ $=\nu*R*\Delta(T) = (P_1-P_2)*V = (1.4-2.2)*10^5*6.8*10^-3 = -949.8 Joules$
the heat is released
$Q = W +U$
$U = Q-W =-949.8 -350 = -1299.8 Joules$