Question

n moles of an ideal monoatomic gas undergo a process in which temp charges with volume as T=KV^2, temp changes from T0 to 4T0 then

Answer 1

ΔU=(9/2) nRT_0, W=(3/2) nRT_0, Q=6 nR*T_0.

Answer 2

Step 1. Relate Volume to Temperature

The process is given by

$$T = K V^2.$$

At the initial state:

$$T_0 = K V_i^2 \quad \Longrightarrow \quad V_i = \sqrt{\frac{T_0}{K}}.$$

At the final state when $T = 4T_0$:

$$4T_0 = K V_f^2 \quad \Longrightarrow \quad V_f = \sqrt{\frac{4T_0}{K}} = 2\sqrt{\frac{T_0}{K}} = 2V_i.$$

Step 2. Change in Internal Energy

For a monatomic ideal gas, the internal energy is

$$U = \frac{3}{2} n R T.$$

Thus, the change in internal energy is

$$\Delta U = \frac{3}{2} n R (T_f - T_i) = \frac{3}{2} n R (4T_0 - T_0) = \frac{3}{2} n R (3T_0) = \frac{9}{2} n R T_0.$$

Step 3. Work Done in the Process

For a quasi–static process, the work done is

$$W = \int_{V_i}^{V_f} P\, dV.$$

Using the ideal gas law $P V = n R T$ and substituting $T=K V^2$, we get

$$P = \frac{n R T}{V} = \frac{n R (K V^2)}{V} = n R K V.$$

Thus,

$$W = \int_{V_i}^{V_f} n R K V\, dV = n R K \left[\frac{V^2}{2}\right]_{V_i}^{V_f}.$$

Substitute $V_f = 2V_i$:

$$W = n R K \left(\frac{(2V_i)^2 - V_i^2}{2}\right) = n R K \left(\frac{4V_i^2 - V_i^2}{2}\right) = n R K \left(\frac{3V_i^2}{2}\right).$$

Since $T_0 = K V_i^2$, we have

$$W = \frac{3}{2} n R T_0.$$

Step 4. Heat Absorbed

Using the first law of thermodynamics,

$$Q = \Delta U + W.$$

Substitute the obtained values:

$$Q = \frac{9}{2} n R T_0 + \frac{3}{2} n R T_0 = 6 n R T_0.$$