Question 169: Thermodynamics & KTG

Question

One mole of an ideal diatomic gas expands from volume $V$ to $2 V$ isothermally at a temperature $27^{\circ} \mathrm{C}$ and does $W$ joule of work. If the gas undergoes same magnitude of expansion adiabatically from $27^{\circ} \mathrm{C}$ doing the same amount of work $W$, then its final temperature will be (close to) $\_\_\_\_$ ${ }^{\circ} \mathrm{C}$.
$$ \left(\log _e 2=0.693\right) $$

Accepted Answer

Case -1 :
Isothermal Expansion $(\Delta \mathrm{T}=0)$
For an ideal gas undergoing isothermal expansion from $V_1$ to $V_2$ :
$$ \mathrm{W}_{\mathrm{iso}}=\mathrm{nRT} \ln \left(\frac{\mathrm{~V}_2}{\mathrm{~V}_1}ight) $$
Given :
$\mathrm{n}=1$ mole
$\mathrm{T}=27^{\circ} \mathrm{C}=300 \mathrm{~K}$
$\frac{\mathrm{V}_2}{\mathrm{~V}_1}=\frac{2 \mathrm{~V}}{\mathrm{~V}}=2$
$\ln 2=0.693$
$$ W=1 	imes R 	imes 300 	imes 0.693=207.9 R $$
Case - 2 :
Adiabatic Expansion ( $\Delta \mathrm{Q}$ )
For an adiabatic process, the work done is given by:
$$ \mathrm{W}_{\mathrm{adia}}=\frac{\mathrm{nR}\left(\mathrm{~T}_1-\mathrm{T}_2ight)}{\gamma-1} $$
The gas is diatomic, so the adiabatic constant $\gamma=\frac{7}{5}=1.4$.
$$ \mathrm{T}_1=300 \mathrm{~K} $$
Work done is the same magnitude ( $\mathrm{W}=207.9 \mathrm{R}$ ).
$$ 207.9 \mathrm{R}=\frac{1 	imes \mathrm{R}\left(300-\mathrm{T}_2ight)}{1.4-1} $$
$$\Rightarrow $$ $207.9=\frac{300-\mathrm{T}_2}{0.4}$
$$\Rightarrow $$ $83.16=300-\mathrm{T}_2$
$$\Rightarrow $$ $\mathrm{T}_2=216.84 \mathrm{~K}$
$$\Rightarrow $$ $\mathrm{T}\left({ }^{\circ} \mathrm{C}ight)=216.84-273.15$
$$ \mathrm{T}\left({ }^{\circ} \mathrm{C}ight) \approx-56.31^{\circ} \mathrm{C} $$
The final temperature is approximately $-56^{\circ} \mathrm{C}$.
Hence, the correct option is (A).

Answer

-56

Answer

-117

Answer

-30

Answer

-189

Thermodynamics & KTG

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