Question

In an industrial process the volume of 25.0 mol of a monatomic ideal gas is reduced at a uniform rate from 0.616 m3 to 0.308 m3 in 2.00 h while its temperature is increased at a uniform rate from 27.0°C to 450°C. Throughout the process, the gas passes through thermodynamic equilibrium states. What are (a) the cumulative work done by the gas, (b) the cumulative energy absorbed by the gas as heat, and (c) the molar specific heat for the process? (Hint: To evaluate the integral for the work, you might use a +bx bx aB- bA In (A Bx), an indefinite integral.) Suppose the process is replaced with a two-step process that reaches the same final state. In step 1, the gas volume is reduced at constant temperature, and in step 2 the temperature is increased at constant volume. For this process, what are (d) the cumulative work done on the gas, (e) the cumulative energy absorbed by the gas as heat, and (f) the molar specific heat for the process? UnitsT UnitsT J UnitsT J/mol-K (a) Numbe (b) Numbe the tolerance is +/-290 (c) Numbe (d) Numbe (e) Numbe (f) Numbe J/mol K Click if you would like to Show Work for this question: Open Show Wor!k

Answer 1

Given,
25. Vị 0.6167723. TI 27"C n
and
V0.308m3 T450
. The process takes place in $\small t=2hrs=7200s$ .

As the volume and temperature both change uniformly, we have  
dV 0.308 -0.616 -7200 -4.2778 * 10-5,n dt
and  
dT dt 450-27 = 0.0588°C/s 7200

Now calculate to get  
d1 dV -1373. 3695°C/ 7n
. Assuming this is the slope of a line equation $\small T=mV+c$ , where $\small m= \frac{dT}{dV}$ and $\small c$ is a constant which can be found by putting $\small (V_1,T_1)$ or
V2.T2
in the equation.

a.) The cumulative work done on the system is

nRT

b.) As $\small U=\frac{3}{2}nRT$ for monoatomic ideal gas, we can find $\small \Delta U=\frac{3}{2}nR(T_2-T_1)$ . Using this and $\small \Delta W$ from above, we can get from 1st Law, the heat energy absorbed $\small \Delta Q=\Delta U-\Delta W$ .

c.) The molar specific heats of ideal monoatomic gases are:

2
and $\small C_P=\frac{1}{n}\left ( \frac{\partial Q}{\partial T} \right )_P=\frac{5}{2}R$

d.) The process is changed to a 2-step process with step-1 being isothermal compression and step-2 being isochoric. The work done in isochoric process is zero because volume is constant. Therefore work done is only due to the isotheramal process.

V2 nRT1

e.) The internal energy depends only on the initial and final temperatures. So it is same as in the previous case$\small \Delta U'=\Delta U$. From 1st Law, the heat energy absorbed $\small \Delta Q'=\Delta U'-\Delta W'$ .

f.) The molar specific heats are also same as the gas is ideal.

Note: I have only given the detailed method to solve the above question and left the calculation part for the reader. This was due to very long time if goes to answering by numerical methods. . But this also helps the students to get an exercise. Hope u understand!