Question No. 5

A gas is enclosed in a cylinder with a movable frictionless piston. Its initial thermodynamic state at pressure $P_i={10}^5\text{Pa}$ and volume $V_i={10}^{-3}{\text{m}}^3$ changes to a final state at $P_f=(1\mathrm{/}32)\times {10}^5\text{Pa}$ and $V_f=8\times {10}^{-3}{\text{m}}^3$ in an adiabatic quasi-static process, such that $P^3{V}^5=\text{constant}$. Consider another thermodynamic process that brings the system from the same initial state to the same final state in two steps: an isobaric expansion at $P_i$ followed by an isochoric (isovolumetric) process at volume $V_f$. The amount of heat supplied to the system in the two-step process is approximately
(A) 112 J
(B) 294 J
(C) 588 J
(D) 813 J

---

Solution:

Step 1: Identify the gas type from the adiabatic process

Given: $P^3{V}^5=\text{constant}$
For an adiabatic process, $P{V}^{\gamma }=\text{constant}$.
Comparing exponents: $\gamma =\frac{5}{3}$
This corresponds to a monatomic gas (e.g., ideal gas with 3 degrees of freedom).
For a monatomic gas:

• $C_v=\frac{3}{2}R$

• $C_p=\frac{5}{2}R$

---

Step 2: Two-step process (isobaric + isochoric)

Step A: Isobaric expansion at $P_i={10}^5\text{Pa}$

• Initial volume: $V_i={10}^{-3}{\text{m}}^3$

• Final volume: $V_f=8\times {10}^{-3}{\text{m}}^3$

• Work done: $W_1=P_i\mathrm{\Delta }V=P_i(V_f-V_i)={10}^5\times (8\times {10}^{-3}-{10}^{-3})={10}^5\times 7\times {10}^{-3}=700\text{J}$

• Change in internal energy: $\mathrm{\Delta }{U}_1=n{C}_v\mathrm{\Delta }T$

Step B: Isochoric process at $V_f$

• Pressure changes from $P_i$ to $P_f=\frac{1}{32}\times {10}^5\text{Pa}$

• Work done: $W_2=0$ (since volume constant)

• Change in internal energy: $\mathrm{\Delta }{U}_2=n{C}_v\mathrm{\Delta }T$

---

Step 3: Compute total change in internal energy ($\mathrm{\Delta }U$)

Since internal energy is a state function, $\mathrm{\Delta }U$ depends only on initial and final states.
For an ideal gas: $\mathrm{\Delta }U=n{C}_v\mathrm{\Delta }T=\frac{3}{2}nR\mathrm{\Delta }T=\frac{3}{2}(P_f{V}_f-P_i{V}_i)$

Calculate $P_i{V}_i$ and $P_f{V}_f$:

• $P_i{V}_i={10}^5\times {10}^{-3}=100\text{J}$

• $P_f{V}_f=(\frac{1}{32}\times {10}^5)\times (8\times {10}^{-3})=\frac{1}{32}\times {10}^5\times 8\times {10}^{-3}=\frac{1}{32}\times 800=25\text{J}$

So,

$$\mathrm{\Delta }U=\frac{3}{2}(P_f{V}_f-P_i{V}_i)=\frac{3}{2}(25-100)=\frac{3}{2}\times (-75)=-112.5\text{J}$$

---

Step 4: Compute total work done ($W$)

• Work in isobaric step: $W_1=700\text{J}$

• Work in isochoric step: $W_2=0$

• Total work: $W=W_1+W_2=700\text{J}$

---

Step 5: Compute heat supplied ($Q$) using first law

$$Q=\mathrm{\Delta }U+W=-112.5+700=587.5\text{J}\approx 588\text{J}$$

---

Final Answer:

$$\boxed{{\displaystyle 588}}$$