Question

You have recorded measurements of the heat flow Q into 0.300 mol of a gas that starts at T₁ = 20.0°C and ends at a temperature T₂. You measured Q for three processes: one isobaric, one isochoric, and one adiabatic. In each case, T₂ was the same. Figure P19.59 summarizes your results. But you lost a page from your lab notebook and don’t have a record of the value of T₂; you also don’t know which process was isobaric, isochoric, or adiabatic. Each process was done at a sufficiently low pressure for the gas to be treated as ideal. (a) Identify each process a, b, or c as isobaric, isochoric, or adiabatic. (b) What is the value of T₂? (c) How much work is done by the gas in each process? (d) For which process is the magnitude of the volume change the greatest? (e) For each process, does the volume of the gas increase, decrease, or stay the same?

Short answer

(a) anadiabatic, b isochoric and cisobaric

(b) the value of T2Final temperature = 28.0$°\mathrm{C}$

(c)thework is done by the gas in each process isWa = -30 J,Wb = 0 and Wc = 20 J

(d) ) For each process, does the volume of the gas decreases for a and for b it remains same and for c it increases.

Step-by-step solution

Step 1

number of moles n = 0.300 mol at specific heat Q and

initial temperature T1=20

three processes have the same final temperature T2.

Step 2:

a)

The heat is given at zero Q =0 i.e. No heat transfer, therefore in process a is an adiabatic process.

For process b and c by the first law of thermodynamics, the heat Q

Q=W+ΔU (1)

U is the same for the both a and b because of, the temperature is the same of the three processes

the volume is constant for the isochoric process the process and asΔU is the same for process b and c therefore, the work done is zero.so, both of them will be the isochoric process,

b has the less heat than c therefore b is isochoric process and

process cis isobaric process

thus,a is adiabatic, b is isochoric and ci s isobaric

b) for )the value of T2Final temperature

take the value of the heat Q for b the isochoric process and W=0

put all the values in equation 1,

$$\begin{gathered} \mathrm{Q}=∆\mathrm{U} \\ \mathrm{Q}={\mathrm{nC}}_{\mathrm{V}}∆\mathrm{T} \\ \mathrm{Q}={\mathrm{nC}}_{\mathrm{V}}\left({\mathrm{T}}_2-{\mathrm{T}}_1\right) \end{gathered}$$

$\mathrm{T}2=\left(\frac{\mathrm{Q}}{{\mathrm{nC}}_{\mathrm{V}}}\right)+\mathrm{T}1$ (2)

CV is the molar heat capacity at constant volume, take CV=3R/2 and R is the gas constant and equals =8.314 J/mol⋅K. put all the values in equation 2,

$$\begin{gathered} \mathrm{T}2=\left(\frac{{\mathrm{Q}}_{\mathrm{b}}}{{\mathrm{nC}}_{\mathrm{V}}}\right)+\mathrm{T}1 \\ \mathrm{T}1=\left(\frac{30\mathrm{J}}{0.300\mathrm{mol}\left({\displaystyle \frac{3\mathrm{R}}{2}}\right)}\right)+293\mathrm{K} \\ \mathrm{T}1=301\mathrm{K}=28.0\mathrm{C}° \end{gathered}$$

Thus, the value of T2 Final temperature = 28.0$°\mathrm{C}$

Step 3

c) thework done by the gas in each process

the gas is insulated and no heat transfer Q =0 and the work done is,

$$\begin{gathered} \mathrm{Wa}=\mathrm{Q}-∆\mathrm{U} \\ =0-{\mathrm{nC}}_{\mathrm{V}}∆\mathrm{T} \\ =-0.300\left(\frac{3\mathrm{R}}{2}\right)\left(301\mathrm{K}-293\mathrm{K}\right) \end{gathered}$$

The work done is negative so the work is done on the gas.

Foe Isochoric process b:

for the isochoric process, the volume is constant and work done is zero

${\mathrm{W}}_{\mathrm{b}}=0$

Isobaric process c

the pressure is constant for the isobaric process, and the work done is,

$\mathrm{W}=\mathrm{p}∆\mathrm{v}$

$\mathrm{W}=\mathrm{pV}2-\mathrm{pV}1$ (3)

the ideal gas law nRT instead of pV where p, ,n and R are constant and T is known,

so,

${\mathrm{W}}_{\mathrm{C}}=\mathrm{nRT}2-\mathrm{nRT}1$ 3^*

Put all the values in 3,

$$\begin{gathered} {\mathrm{W}}_{\mathrm{C}}=\mathrm{nRT}\left(\mathrm{T}2-\mathrm{T}1\right) \\ {\mathrm{W}}_{\mathrm{C}}=0.300\times 8.314\times \left(301\mathrm{K}-293\mathrm{K}\right) \\ {\mathrm{W}}_{\mathrm{C}}=20\mathrm{J} \end{gathered}$$

The work done is positive so the work is done by the gas.

Thus,thework is done by the gas in each process isWa = -30 J,Wb = 0 and Wc = 20 J

Step 4

d)For each process, does the volume of the gas increase or decrease.

The work done depends on the change in volume, work done$\left|\mathrm{W}\right|=\left|-30\right|\mathrm{J}=30\mathrm{J}$ is due to the great change in volume.

the greatest magnitude of the work donewhich isadiabatic process and there is the greatest change in volume.

from (c) determine for each process,

In Adiabatic process the work done is negative, and so the work is done on the gas where gas is compressed as a result the volume decrease.

In Isochoric process the volume is constant and the volume stay the same. Also, the work done is positive and the work is done by the gas and the gas expands, as a result the volume increase.

Thus, for each process, does the volume of the gas decreases for a and for b it remains same and for c it increases.