© All Rights reserved @ LearnWithDash
Step-by-Step Solution
Step 1: Identify the Given Information
• Amount of gas: 1 kilo mole (which is 1000 moles).
• Gas constant, $R = 8.3\, \text{J mol}^{-1}\text{K}^{-1}$.
• Work done on the gas adiabatically: $W = -146\,\text{kJ} = -146000\,\text{J}$ (negative sign indicates work done on the gas).
• Rise in temperature: $\Delta T = 7\,^\circ\text{C} = 7\,\text{K}$ (a change in temperature is the same in Celsius and Kelvin).
Step 2: Write the Relevant Equation for Adiabatic Process
For an adiabatic process, the work done on the gas can be related to the change in temperature using the following formula:
$W = \frac{n R\, \Delta T}{1 - \gamma}\,,$
where $n$ is the number of moles of the gas, $R$ is the gas constant, $\Delta T$ is the temperature change, and $\gamma$ (gamma) is the ratio of specific heats $\left(\gamma = \frac{C_p}{C_v}\right)$.
Step 3: Substitute the Known Values and Solve for $ \gamma $
Substituting the given values into the formula:
$-146000 = \frac{(1000)\times (8.3)\times (7)}{1 - \gamma} \,.$
Simplify the right-hand side:
$\frac{(1000)\times (8.3)\times (7)}{1 - \gamma} = \frac{58100}{1 - \gamma}\,.$
Thus, we have:
$-146000 = \frac{58100}{1 - \gamma}\,.$
Rearrange to solve for $1 - \gamma$:
$1 - \gamma = -\frac{58100}{146000} = -\frac{58.1}{146} \,.$
Now compute:
$1 - \gamma = -0.398 \quad \Longrightarrow \quad \gamma = 1.398 \approx 1.4 \,.$
Step 4: Interpret the Value of $ \gamma $
The ratio of specific heats $ \gamma = 1.4 $ matches closely to that of a diatomic gas. Therefore, the gas is diatomic.
