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

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Quantrex Academy · Accepted Answer

Correct answer: C. 0.9

According to the principle of calorimetry, the heat gained by the gas is equal to the heat released by the condensation of steam.

m_{\text{gas}} c_v \Delta T = m_{\text{steam}} L

Substituting the given values:

3 \times c_v \times (100 - 20) = 0.095 \times 540

3 \times c_v \times 80 = 51.3

240 c_v = 51.3

c_v = \dfrac{51.3}{240} = 0.21375 \text{ cal g}^{-1} \text{ K}^{-1}

To convert the specific heat capacity into

\text{J g}^{-1} \text{ K}^{-1}

, multiply by the mechanical equivalent of heat (

1 \text{ cal} \approx 4.2 \text{ J}

):

c_v = 0.21375 \times 4.2 = 0.89775 \text{ J g}^{-1} \text{ K}^{-1}

Rounding to one decimal place gives

0.9 \text{ J g}^{-1} \text{ K}^{-1}

.

Within Joly's differential steam calorimeter, 3 g of an ideal gas...

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