One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27°C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27°C doing the same amount of work W, then its final temperature will be (close to) __

One mole of an ideal diatomic gas expands from volume V to 2V isothermally at a temperature 27°C and does W joule of work. If the gas undergoes same magnitude of expansion adiabatically from 27°C doing the same amount of work W, then its final temperature will be (close to) ___ °C.

Q. One mole of an ideal diatomic gas expands from volume $V$ to $2V$ isothermally at a temperature $27^\circ C$ and does $W$ joule of work. If the gas undergoes same magnitude of expansion adiabatically from $27^\circ C$ doing the same amount of work $W$, then its final temperature will be (close to) ___ $^\circ C$.

$(\log_e 2 = 0.693)$

A. $-56$

B. $-117$

C. $-30$

Correct Answer: −56

Explanation

For isothermal expansion of one mole of an ideal gas, the work done is

$$ W = nRT \ln \frac{V_2}{V_1} $$

Here $n=1$, $T = 27^\circ C = 300\,K$ and $\dfrac{V_2}{V_1} = 2$.

So, $$ W = R \cdot 300 \cdot \ln 2 $$

For adiabatic expansion, work done is equal to decrease in internal energy:

$$ W = C_V (T_1 - T_2) $$

For a diatomic gas, $$ C_V = \frac{5}{2}R $$

Thus, $$ R \cdot 300 \cdot \ln 2 = \frac{5}{2}R (300 - T_2) $$

Cancelling $R$, $$ 300 \ln 2 = \frac{5}{2}(300 - T_2) $$

Using $\ln 2 = 0.693$, $$ 300 \times 0.693 = \frac{5}{2}(300 - T_2) $$

$$ 207.9 = \frac{5}{2}(300 - T_2) $$

$$ 300 - T_2 = \frac{2}{5} \times 207.9 \approx 83.2 $$

$$ T_2 \approx 216.8\,K $$

Converting to Celsius: $$ 216.8 - 273 \approx -56^\circ C $$

Explanation

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