(Take the universal gas constant as $R = 8.3\text{ J mol}^{-1}\text{ K}^{-1}$)
For an adiabatic process involving an ideal gas, the work done ($W$) is related to the change in internal energy ($\Delta U$). According to the first law of thermodynamics, $\Delta U = Q - W$. Since the process is adiabatic, $Q = 0$, hence $\Delta U = -W$.
The change in internal energy for $n$ moles of an ideal gas is given by:
$\Delta U = n C_V \Delta T$
Where $C_V$ is the molar specific heat at constant volume and $\Delta T$ is the change in temperature. For a monatomic ideal gas, $C_V = \frac{R}{\gamma - 1}$.
Combining these, the work done by the gas is:
$W = -\Delta U = -n C_V \Delta T = -n \left(\frac{R}{\gamma - 1}\right) (T_2 - T_1) = n \left(\frac{R}{\gamma - 1}\right) (T_1 - T_2)$
$\gamma - 1 = \frac{5}{3} - 1 = \frac{5 - 3}{3} = \frac{2}{3}$
$T_1 - T_2 = 60 \text{ K} - 50 \text{ K} = 10 \text{ K}$
$W = n \left(\frac{R}{\gamma - 1}\right) (T_1 - T_2)$
$W = (1 \text{ mol}) \left(\frac{8.3 \text{ J mol}^{-1} \text{ K}^{-1}}{2/3}\right) (10 \text{ K})$
$W = 1 \times \left(8.3 \times \frac{3}{2}\right) \times 10 \text{ J}$
$W = \frac{3}{2} \times 8.3 \times 10 \text{ J}$
$W = 1.5 \times 8.3 \times 10 \text{ J}$
$W = 12.45 \times 10 \text{ J}$
$W = 124.5 \text{ J}$
A flask contains argon and chlorine in the ratio of $2:1$ by mass. The temperature of the mixture is $27^\circ\text{C}$. The ratio of root mean square speed of the molecules of the two gases $(\frac{V_{rms}^{Ar}}{V_{rms}^{Cl}})$ is :
(Atomic mass of argon = $40 \text{ u}$ and molecular mass of chlorine = $70 \text{ u}$)
One mole of an ideal monatomic gas undergoes a cyclic process as shown in the figure. The total heat supplied to the gas is :
Rods x and y of equal dimensions but of different materials are joined as shown in figure. Temperatures of end points $A$ and $F$ are maintained at $100^\circ\text{C}$ and $40^\circ\text{C}$ respectively. Given the thermal conductivity of rod x is three times of that of rod y, the temperature at junction points $B$ and $E$ are (close to):
