Verified Solution • 2025–26 Session moderate

QPages 77-80: Q15: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q15 Solution: In Joly's differential steam calorimeter, 3 g of an ideal gas…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often plug the latent heat $L=540\,$cal g⁻¹ directly into the formula for $C_v$ in J g⁻¹ K⁻¹, forgetting the required conversion $1\,$cal = $4.2\,$J; this yields a value about 4.2 times too low. Always convert all energy quantities to the same unit system before computing the specific heat.

In Joly's differential steam calorimeter, 3 g of an ideal gas is contained in a rigid closed sphere at 20�C. The sphere is heated by steam at 100�C and it is found that an extra 0.095 g of steam has condensed into water as the temperature of the gas becomes constant. Calculate the specific heat capacity of the gas in J g−1 K−1\text{J g}^{-1} \text{ K}^{-1}. The latent heat of vaporisation of water = 540 cal g−1540 \text{ cal g}^{-1}.

Step-by-Step Verified Solution

**Answer:** Cv=0.9C_v = 0.9 J/g-K. **Step-by-Step Explanation:** In Joly's differential steam calorimeter: Cv=m2Lm1(θ2−θ1)C_v = \frac{m_2 L}{m_1(\theta_2 - \theta_1)}. Given: m1=3m_1 = 3 g, m2=0.095m_2 = 0.095 g, L=540L = 540 cal/g, θ1=20�C\theta_1 = 20�\text{C}, θ2=100�C\theta_2 = 100�\text{C}.
Cv=0.095×540×4.23×80=0.89≈0.9 J/g-KC_v = \frac{0.095 \times 540 \times 4.2}{3 \times 80} = 0.89 \approx 0.9 \text{ J/g-K}

All Questions in this Chapter

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