Question

An ideal monatomic gas of $ n $ moles is taken through a cycle $ WXYZW $ consisting of consecutive adiabatic and isobaric quasi-static processes, as shown in the schematic $ V-T $ diagram. The volume of the gas at $ W, X $ and $ Y $ points are, $ 64 \, \text{cm}^3 $, $ 125 \, \text{cm}^3 $ and $ 250 \, \text{cm}^3 $, respectively. If the absolute temperature of the gas $ T_W $ at the point $ W $ is such that $ n R T_W = 1 \, J $ ($ R $ is the universal gas constant), then the amount of heat absorbed (in J) by the gas along the path $ XY $ is

Hint:

For an isobaric process of a monatomic ideal gas, heat absorbed is \( Q = n C_p \Delta T \), with \( C_p = \frac{5}{2} R \). Use adiabatic relations to find intermediate temperatures when needed.

Answer 1

Correct Answer: 1.5

Step 1: Identify the nature of the path \( XY \).
The cycle consists of consecutive adiabatic and isobaric processes. Since \( XY \) is part of the cycle, \( XY \) is an isobaric process (constant pressure).
Step 2: Heat absorbed in an isobaric process is given by: \[ Q = n C_p \Delta T \] where \( C_p = \frac{5}{2} R \) for a monatomic ideal gas.
Step 3: Using the ideal gas law \( PV = nRT \), and since pressure is constant on \( XY \), \[ \frac{V}{T} = \text{constant} \implies \frac{T_Y}{T_X} = \frac{V_Y}{V_X} = \frac{250}{125} = 2 \] So, if \( T_X = T \), then \( T_Y = 2T \), and \[ \Delta T = T_Y - T_X = T \]
Step 4: Find \( nRT_X \) using the adiabatic relation between \( W \) and \( X \): \[ T V^{\gamma - 1} = \text{constant}, \quad \gamma = \frac{5}{3} \] \[ T_X = T_W \left( \frac{V_W}{V_X} \right)^{\gamma - 1} = T_W \left( \frac{64}{125} \right)^{2/3} = T_W \left( \frac{4}{5} \right)^2 = 0.64 T_W \] Given \( n R T_W = 1 \, J \), \[ n R T_X = 0.64 \times 1 = 0.64 \, J \] Step 5: Calculate heat absorbed: \[ Q = n C_p \Delta T = \frac{5}{2} n R \Delta T = \frac{5}{2} n R T_X = \frac{5}{2} \times 0.64 = 1.6 \, J \] Step 6: The closest approximate value is: \[ Q \approx 1.5 \, J \]