Question

One mole of an ideal monoatomic gas, initially at temperature 𝑇_𝑜 is expanded from an initial volume 𝑉_𝑜 to 2.5𝑉_𝑜. Which of the following statements is(are) correct?
(R is the ideal gas constant)

• A. When the process is isothermal, the work done is 𝑅𝑇_𝑜 ln 2
• B. When the process is isothermal, the change in internal energy is zero
• C. When the process is isobaric, the work done is \(\frac{3}{2}\) RT₀
• D. When the process is isobaric, the change in internal energy is \(\frac{9}{2}\) RT₀

Answer 1

The Correct Option is B, C

1. Isothermal Process 

In an isothermal process, the temperature remains constant, so the internal energy of an ideal gas does not change:

ΔU = 0

The work done in an isothermal expansion is given by:

W = nRT ln(V_f / V_i)

where V_f = 2.5V₀ and V_i = V₀. Substituting:

W = RT₀ ln(2.5)

The expression RT₀ ln(2) is incorrect; hence, statement (A) is not valid.

2. Change in Internal Energy in an Isothermal Process

Since ΔU = 0, statement (B) is correct.

3. Isobaric Process

In an isobaric process, the pressure remains constant. The work done is:

W = PΔV

Using the ideal gas law (PV = nRT), the work done can be expressed as:

W = nRΔT

For a monoatomic ideal gas, ΔT = T_f − T₀, but the given expression (3/2)RT₀ is incorrect.

Thus, statement (C) is invalid.

4. Change in Internal Energy in an Isobaric Process

The change in internal energy is:

ΔU = (3/2) nRΔT

The expression (9/2)RT₀ does not align with this derivation.

Thus, statement (D) is incorrect.

Conclusion

The correct statements are (B) and (C).