Question

Four moles of a diatomic gas at a temperature of 927 $^\circ$C expands adiabatically such that its volume increases by 3100%. The work done by the gas is (Universal gas constant = 8.3 J mol$^{-1}$ K$^{-1}$)

• A. 24.9 kJ
• B. 49.8 kJ
• C. 57.6 kJ
• D. 74.7 kJ

Hint:

For adiabatic processes, remember the relations $T V^{\gamma-1} = \text{constant}$ and $P V^{\gamma} = \text{constant}$. The work done can be directly calculated if initial and final temperatures are known. Always convert temperatures to Kelvin.

Answer 1

The Correct Option is D

Step 1: Convert initial temperature to Kelvin and determine the adiabatic index ($\gamma$).
Initial temperature, $T_1 = 927^\circ \text{C} = 927 + 273 = 1200 \text{ K}$.
For a diatomic gas, the number of degrees of freedom is $f=5$. The adiabatic index ($\gamma$) is given by:
\[ \gamma = 1 + \frac{2}{f} = 1 + \frac{2}{5} = \frac{7}{5} = 1.4 \] Step 2: Determine the relationship between initial and final volumes.
The problem states that the volume increases by 3100%. This means the final volume ($V_2$) is the initial volume ($V_1$) plus 3100% of $V_1$. Percentage increase means: $V_2 = V_1 + \left(\frac{3100}{100}\right) V_1$ $V_2 = V_1 + 31 V_1$ $V_2 = 32 V_1$ So, the ratio $\frac{V_1}{V_2} = \frac{1}{32}$. Step 3: Calculate the final temperature ($T_2$) using the adiabatic relation. For an adiabatic process, the relation between temperature and volume is $T V^{\gamma-1} = \text{constant}$. \[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \] \[ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{\gamma-1} \] Substitute the values: $T_1 = 1200 \text{ K}$, $\frac{V_1}{V_2} = \frac{1}{32}$, and $\gamma - 1 = 1.4 - 1 = 0.4$. \[ T_2 = 1200 \text{ K} \times \left(\frac{1}{32}\right)^{0.4} \] Since $0.4 = \frac{2}{5}$ and $32 = 2^5$: \[ T_2 = 1200 \text{ K} \times \left(\frac{1}{2^5}\right)^{2/5} \] \[ T_2 = 1200 \text{ K} \times (2^{-5})^{2/5} \] \[ T_2 = 1200 \text{ K} \times 2^{(-5 \times 2/5)} \] \[ T_2 = 1200 \text{ K} \times 2^{-2} \] \[ T_2 = 1200 \text{ K} \times \frac{1}{4} \] \[ T_2 = 300 \text{ K} \] Step 4: Calculate the work done by the gas ($W$).
The work done by the gas in an adiabatic process is given by: \[ W = \frac{nR(T_1 - T_2)}{\gamma - 1} \] Substitute the values: $n = 4 \text{ mol}$, $R = 8.3 \text{ J mol}^{-1} \text{ K}^{-1}$, $T_1 = 1200 \text{ K}$, $T_2 = 300 \text{ K}$, and $\gamma - 1 = 0.4$. \[ W = \frac{4 \times 8.3 \times (1200 - 300)}{0.4} \] \[ W = \frac{4 \times 8.3 \times 900}{0.4} \] \[ W = \frac{33.2 \times 900}{0.4} \] To simplify, multiply numerator and denominator by 10: \[ W = \frac{332 \times 900}{4} \] \[ W = 83 \times 900 \] \[ W = 74700 \text{ J} \] Convert Joules to kilojoules: \[ W = \frac{74700}{1000} \text{ kJ} = 74.7 \text{ kJ} \] The final answer is $\boxed{74.7 \text{ kJ}}$.