Rigid-vessel condensation calculation:\n1 kg saturated steam at 100°C (V = 1.673 m^3) is cooled in a rigid vessel to 98°C. Given v_g at 98°C = 1.789 m^3/kg (v_f negligible). How many kilograms of vapor condense?

Introduction / Context:
Steam in a rigid container undergoes condensation upon cooling until the mixture satisfies saturation at the final temperature. This is a common transient in thermal equipment cool-down and storage vessels.

Given Data / Assumptions:

• Initial: 1 kg saturated vapor at 100°C; total volume V_total = 1.673 m^3.
• Final: 98°C saturated mixture in the same rigid volume.
• Specific volume of saturated vapor at 98°C: v_g = 1.789 m^3/kg; specific liquid volume v_f ≈ 0 (negligible compared with v_g).

Concept / Approach:
At equilibrium, total volume equals vapor mass times vapor specific volume plus negligible liquid volume. Thus, m_v = V_total / v_g, and condensed mass m_condensed = m_initial − m_v.

Step-by-Step Solution:
Compute vapor mass at 98°C: m_v = 1.673 / 1.789 ≈ 0.9359 kg.Condensation: m_condensed = 1.000 − 0.9359 ≈ 0.0641 kg.Rounded to the options, m_condensed ≈ 0.065 kg.

Verification / Alternative check:
Because v_g at 98°C is larger than at 100°C, less vapor is needed to occupy the fixed volume; thus some vapor must condense, giving a small positive condensed mass—consistent with the result.

Why Other Options Are Wrong:
0.0 implies no condensation (incorrect); 0.1 is too large; 1.0 means complete condensation; 0.33 is arbitrary and inconsistent with the volume ratio.

Common Pitfalls:
Forgetting to use absolute total vessel volume; neglecting that v_f is negligible but not negative; mixing specific volume data at different temperatures.

Final Answer:
0.065