In Joly's differential steam calorimeter, `3g` of an ideal gas is ccontained 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 `(Jg^(-1) K^(-1).` The latent heat of vaporization of `water = 540 cal g ^(-1)`.

Explanation

To find the specific heat capacity of the gas using Joly's differential steam calorimeter, we can follow these steps:

### Step 1: Understand the problem
We have 3 g of an ideal gas at an initial temperature of 20 °C. The gas is heated by steam at 100 °C, and during this process, 0.095 g of steam condenses into water. We need to calculate the specific heat capacity of the gas.

### Step 2: Identify the relevant formulas
The heat gained by the gas (Q_gas) is equal to the heat lost by the steam (Q_steam). The heat gained by the gas can be expressed as:
\[ Q_{\text{gas}} = m_1 \cdot C_v \cdot (T_2 - T_1) \]
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