91Ó°ÊÓ

If a system after passing through a sequence of states, comes back to the state from which it started, then the process is known as a cyclic process. In a cyclic process the internal energy of the system remains conserved.

• Number of moles of an ideal monoatomic gas,n=1.00mol
• Temperatures at points 1,2and 3 are$T_1=300 K,T_2=600 K,and T_3=455K$ respectively.

Initial pressure at point 1 is$P_1=1.0atm$.

$Q=C_{v}n\mathrm{Δ}T$

But,

$C_v=\frac{3}{2}R, (R=Gasconstant)$

So,

$Q=\frac{3}{2}Rn\mathrm{Δ}T$

$Q=\frac{3}{2}(1.00 mol)(8.31 \frac{J}{mol}.K)(600 K-300 K)$

$Q=3.74×{10}^{3}J$

Therefore,the heat Q for$1→2$is$3.74×{10}^3 J$

According to first law of thermodynamics,

$Q=W+\mathrm{Δ}{E}_{int}$

But, work done at constant volume is zero. So,

W=0J

Therefore,

$$\begin{gathered} Q=\mathrm{Δ}{E}_{int} \\ \mathrm{Δ}{E}_{int}=3,74×{10}^{3}J \end{gathered}$$

Therefore, the change in internal energy $\mathrm{Δ}{E}_{int}$for$1→2$is$3.74×{10}^3 J$.

c)

As volume is constant for$1→2$and work done for a thermodynamic process is given as-

$W=\mathrm{Î\pounds }\left(P\mathrm{Δ}V\right)$

So, for $\mathrm{Δ}V=0$, we have

W=P$×$0

=0

Therefore,work done at constant volume is zero.

For$2→3$,

d)

As the process from $2→3$is adiabatic, no heat is transferred.

Therefore, Q=0J.

e)

Using first law of thermodynamics,

$$\begin{gathered} Q=W+\mathrm{Δ}{E}_{int} \\ 0=W+\mathrm{Δ}{E}_{int} \end{gathered}$$

$W=-\mathrm{Δ}{E}_{int}$

For the equation,

$W=\frac{nR\mathrm{Δ}T}{(1-\mathrm{γ})}$

Here, $\mathrm{γ}$is the ratio of specific heats at constant pressure and volume. For a ideal monoatomic gas $\mathrm{γ}=1.66$.

For the given values above equation becomes-

$\mathrm{Δ}{E}_{int}=-\frac{nR\mathrm{Δ}T}{1-\mathrm{γ}}$
$$\begin{gathered} =-\frac{(1.00mol)(8.31{\displaystyle \frac{J}{mol}}.K)(455K-600K)}{1-(1.66)} \\ =-1.82×{10}^{3}J \end{gathered}$$

Therefore, change in internal energy $\mathrm{Δ}{E}_{int}$for $2→3$is $-1.82×{10}^3 J$.

According to the first law of thermodynamics,

$Q=W+\mathrm{Δ}{E}_{int}$

But, no heat added for adiabatic process. So,

Q=0J

Therefore,

$$\begin{gathered} 0=W+\mathrm{Δ}{E}_{int} \\ W=-\mathrm{Δ}{E}_{int} \\ W=-(-1.82×{10}^3 J) \\ W=1.82×{10}^{3}J \end{gathered}$$

Therefore, the work done W for $2→3$is$1.82×{10}^{3}J$

For $3→2$,

For constant pressure,

$C_p=\frac{Q}{n\mathrm{Δ}T}$
$Q=C_{P}n\mathrm{Δ}T$

But,

localid="1662540584797" $C_P=\frac{5}{2}R$

So,$Q=\frac{5}{2}Rn\mathrm{Δ}T$

For the given values, the above equation becomes-

localid="1662479313288" $Q=\frac{5}{2}(1.00mol)(8.31 \frac{J}{mol}.K)( 300 K-455 K)$

$Q=-3.22×{10}^3 J$

Therefore, the heat Q for$3→1$is$-3.22×{10}^{3}J$

h)

We have

$C_v=\frac{\left(\mathrm{Δ}{E}_{i}nt\right)}{n\mathrm{Δ}T}$

But,

$\mathrm{Δ}{E}_{int}=\frac{3}{2}(8.31\frac{J}{mol}.K)(1.00 mol)(300 K-455 K)$

$\mathrm{Δ}{E}_{i}nt=-1.93×{10}^3 J$

Therefore, change in internal energy $\mathrm{Δ}{E}_{int}$for$3→1$ is -1,93*10³J.

i)

According to first law of thermodynamics,

$Q=W+\mathrm{Δ}{E}_{int}$

But,$Q=-3.22×{10}^3 J$and$\mathrm{Δ}{E}_{i}nt=-1.93×{10}^3 J$

localid="1662480339693" $\begin{array}{l}=-1.29×{10}^3]\end{array}$

Therefore, the work done w for 3_1 is -1.29*10³J

For the full cycle-

$Q=3.74×{10}^3 J+0 J-3.{2210}^3 J$

Q=520J

Therefore,heat for full cycle is 520J.

For the full cycle, total change in internal energy is

$\mathrm{Δ}{E}_{int}=3.74×{10}^{3}J-1.81×{10}^{3}J-1.93×{10}^{3}J$

$\mathrm{Δ}{E}_{int}=0J$

Therefore, the change in internal energy$\mathrm{Δ}{E}_{int}$for full cycle is0J .

For full cycle, the total work done is

$W=0 J+1.81×{10}^3 J-1.29×{10}^3 J$

W=520J

Therefore, the work done W for full cycle is 520J.

m)

By ideal gas equation,

$P_1{V}_1=nR{T}_1$

$$\begin{gathered} V_1=\frac{nR{T}_1}{P_1} \\ =\frac{(1.00mol)\left(8.31{\displaystyle \frac{J}{mol}}.K\right)\left(300K\right)}{1.013×{10}^{5}Pa} \\ =2.46×{10}^{-2}{m}^3 \end{gathered}$$

Therefore, volume at point 2 is$2.46×{10}^{-2}{m}^3$.

An ideal gas equation,

$PV=nRT$

As $1→2$process is an isochoric process,V₁=V₂

$$\begin{gathered} P_2=\frac{nR{T}_1}{V_2} \\ =\frac{(1.00mol)\left(8.31{\displaystyle \frac{J}{mol}}.K\right)\left(600K\right)}{2.46×{10}^{-2}{m}^3} \\ =2.0×{10}^{5}Pa \\ =2atm \end{gathered}$$

Therefore, pressure at point 2 is 2atm.

o)

For the ideal gas equation

PV=nRT

So, volume for state 3 is given as,

$$\begin{gathered} V_3=\frac{nR{T}_3}{P_3} \\ =\frac{(1.00mol)\left(8.31{\displaystyle \frac{J}{mol}}.K\right)\left(455K\right)}{1.013×{10}^{5}Pa} \\ 3.73×{10}^{-2}{m}^3 \end{gathered}$$

Therefore, volume at point 3 is $3.73×{10}^{-2}{m}^3$.

p)

PV=nRT

Pressure of state 1 and 3 is same. So,

$$\begin{gathered} P_3=1.013×{10}^5\mathrm{Pa} \\ =1atm \end{gathered}$$

Therefore, pressure at point 3 isrole="math" localid="1662616288722" $1.013×{10}^{5}Pa$.