Question

one mole of an ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°c. if the work done during the process is 3kj, the final temperature will be equal to ($c_{v}=20j/k$)

Explanation:

Step1: Recall adiabatic work - energy relation

For an adiabatic process of an ideal gas, $W = nC_V\Delta T$. Here, $n = 1$ mole, $C_V=20\ J/K$, and $W = 3000\ J$.

Step2: Rearrange the formula for $\Delta T$

$\Delta T=\frac{W}{nC_V}$. Substituting the values: $\Delta T=\frac{3000\ J}{1\ mol\times20\ J/K}=150\ K$.

Step3: Calculate the final temperature

The initial temperature $T_1=(27 + 273)K=300\ K$. Since work is done by the gas, the internal energy decreases and the temperature drops. So, $T_2=T_1-\Delta T$.
$T_2 = 300\ K-150\ K = 150\ K$.

Answer:

$150\ K$