Within Joly's differential steam calorimeter, 3 g of an ideal gas resides in a rigid closed sphere at 20^ C . The sphere is heated using steam at 100^ C . Once the gas temperature becomes constant, it is observed that an additional 0.095 g of steam has condensed into water. Determine the specific heat capacity of the gas in J g ⁻¹ K ⁻¹ . The latent heat of vaporization of water is 540 cal g ⁻¹ .

Step-by-step solution

According to the principle of calorimetry, the heat gained by the gas is equal to the heat released by the condensation of steam. m_ gas c_v T = m_ steam L Substituting the given values: 3 c_v (100 - 20) = 0.095 540 3 c_v 80 = 51.3 240 c_v = 51.3 c_v = 51.3 240 = 0.21375 cal g ⁻¹ K ⁻¹ To convert the specific heat capacity into J g ⁻¹ K ⁻¹ , multiply by the mechanical equivalent of heat ( 1 cal 4.2 J ): c_v = 0.21375 4.2 = 0.89775 J g ⁻¹ K ⁻¹ Rounding to one decimal place gives 0.9 J g ⁻¹ K ⁻¹ .