Question

in a quasistatic process, an ideal monoatomic gas is applied heat in sch a way that volincreases an freq of collisions of atoms remains constant if molar specific heat of gas is nR find n

Answer 1

2

Answer 2

Solution:

1. Collision Frequency Constraint:

For an ideal gas, the collision frequency is proportional to the product of the number density and the average speed:

$$f \propto \frac{1}{V}\sqrt{T}$$

Keeping $f$ constant implies

$$\frac{\sqrt{T}}{V} = \text{constant} \quad \Longrightarrow \quad T \propto V^2.$$

Write

$$T = aV^2 \quad \text{with } a \text{ constant.}$$

2. First Law of Thermodynamics:

For one mole of a monatomic ideal gas, the first law is

$$dQ = dU + PdV,$$

and

$$dU = \frac{3}{2}R\,dT, \quad P = \frac{RT}{V}.$$

Thus,

$$dQ = \frac{3}{2}R\,dT + \frac{RT}{V}\,dV.$$

3. Relating $dT$ and $dV$:

Since $T = aV^2$, differentiating gives:

$$dT = 2aV\,dV.$$

Also, note that

$$\frac{T}{V} = aV.$$

4. Substitute in the First Law:

$$dQ = \frac{3}{2}R (2aV\,dV) + R(aV) \,dV = 3RaV\,dV + RaV\,dV = 4RaV\,dV.$$

Express $dQ$ in terms of $dT$:

$$dT = 2aV\,dV \quad \Longrightarrow \quad dV = \frac{dT}{2aV}.$$

Substitute:

$$dQ = 4RaV \cdot \frac{dT}{2aV} = 2R\,dT.$$

5. Determine the Molar Specific Heat:

By definition, for this process,

$$dQ = C\,dT \quad \text{with } C = nR.$$

Comparing, we have:

$$nR = 2R \quad \Longrightarrow \quad n = 2.$$

Explanation (Minimal):

• Collision frequency constant implies $\sqrt{T}/V$ is constant $\Rightarrow T \propto V^2$.
• Use $T = aV^2$ to relate $dT$ and $dV$.
• Apply first law: $dQ = \tfrac{3}{2}RdT + (RT/V)dV$.
• Substitute $dT = 2aV\,dV$ and simplify to get $dQ = 2R\,dT$.
• Thus, molar specific heat $C = 2R$ i.e. $n = 2$.