Question

A vertical cylinder contains an ideal gas at atmospheric pressure. The cylinder has a 5kg movable piston at the bottom which has a surface area $5 \times 10^{-3}m^2$. Now, the gas is heated slowly from 300K to 350K and the piston rises by 10cm. The piston is now clamped at this position and the gas is now cooled back to 300K. Find the difference between the heat provided during the heating process and the heat lost in the cooling process.

• A. 5 J
• B. 50 J
• C. 55 J
• D. None of these

Answer 1

Answer: (C)

55 J

Answer 2

The problem involves two processes for an ideal gas: an isobaric heating process followed by an isochoric cooling process. We need to find the difference between the heat provided during heating and the heat lost during cooling.

Given Data:

• Mass of piston, $m = 5 \, \text{kg}$
• Surface area of piston, $A = 5 \times 10^{-3} \, \text{m}^2$
• Initial temperature, $T_1 = 300 \, \text{K}$
• Final temperature after heating, $T_2 = 350 \, \text{K}$
• Rise in piston, $\Delta h = 10 \, \text{cm} = 0.1 \, \text{m}$
• Atmospheric pressure, $P_0 = 10^5 \, \text{Pa}$
• Acceleration due to gravity, $g = 9.8 \, \text{m/s}^2$

Process 1: Heating (Isobaric Process)

During heating, the piston is movable, so the pressure inside the cylinder remains constant. This constant pressure is the sum of atmospheric pressure and the pressure due to the piston's weight.

1. Calculate the total pressure ($P_{total}$):

   $P_{total} = P_0 + \frac{mg}{A}$

   $P_{total} = 10^5 \, \text{Pa} + \frac{5 \, \text{kg} \times 9.8 \, \text{m/s}^2}{5 \times 10^{-3} \, \text{m}^2}$

   $P_{total} = 10^5 + \frac{49}{5 \times 10^{-3}} = 10^5 + 9800 = 109800 \, \text{Pa}$

2. Calculate the change in volume ($\Delta V_1$):

   $\Delta V_1 = A \times \Delta h$

   $\Delta V_1 = (5 \times 10^{-3} \, \text{m}^2) \times (0.1 \, \text{m}) = 5 \times 10^{-4} \, \text{m}^3$

3. Calculate the work done by the gas ($W_1$):

   Since it's an isobaric process, $W_1 = P_{total} \Delta V_1$

   $W_1 = 109800 \, \text{Pa} \times 5 \times 10^{-4} \, \text{m}^3 = 54.9 \, \text{J}$

4. Apply the First Law of Thermodynamics for Process 1:

   $\Delta Q_1 = \Delta U_1 + W_1$

   Where $\Delta U_1 = nC_v(T_2 - T_1)$ is the change in internal energy.

   $\Delta Q_1 = nC_v(350 \, \text{K} - 300 \, \text{K}) + 54.9 \, \text{J}$

   $\Delta Q_1 = nC_v(50 \, \text{K}) + 54.9 \, \text{J}$

Process 2: Cooling (Isochoric Process)

The piston is clamped, so the volume remains constant. The gas is cooled from $T_2 = 350 \, \text{K}$ back to $T_1 = 300 \, \text{K}$.

1. Calculate the work done by the gas ($W_2$):

   Since the volume is constant, $\Delta V_2 = 0$, so $W_2 = P \Delta V_2 = 0$.

2. Apply the First Law of Thermodynamics for Process 2:

   $\Delta Q_2 = \Delta U_2 + W_2$

   $\Delta Q_2 = \Delta U_2 + 0 = \Delta U_2$

   Where $\Delta U_2 = nC_v(T_1 - T_2)$ is the change in internal energy.

   $\Delta Q_2 = nC_v(300 \, \text{K} - 350 \, \text{K})$

   $\Delta Q_2 = nC_v(-50 \, \text{K})$

Finding the Difference between Heat Provided and Heat Lost:

Heat provided during heating = $\Delta Q_1$

Heat lost in cooling process = $-\Delta Q_2$ (since $\Delta Q_2$ is negative, heat lost is a positive value)

Difference $= \Delta Q_1 - (-\Delta Q_2)$

Difference $= \Delta Q_1 + \Delta Q_2$

Alternatively, consider the entire process from $T_1$ to $T_2$ and back to $T_1$. The net change in internal energy is zero because the gas returns to its initial temperature:

$\Delta U_{net} = \Delta U_1 + \Delta U_2 = nC_v(T_2 - T_1) + nC_v(T_1 - T_2) = 0$.

From the First Law of Thermodynamics for the entire process:

$\Delta Q_{net} = \Delta U_{net} + W_{net}$

$\Delta Q_1 + \Delta Q_2 = 0 + (W_1 + W_2)$

$\Delta Q_1 + \Delta Q_2 = W_1 + 0$

$\Delta Q_1 + \Delta Q_2 = W_1$

The question asks for the difference between the heat provided ($\Delta Q_1$) and the heat lost ($-\Delta Q_2$).

So, the required difference is $\Delta Q_1 - (-\Delta Q_2) = \Delta Q_1 + \Delta Q_2$.

As derived above, this is equal to $W_1$.

Difference $= W_1 = 54.9 \, \text{J}$

Rounding to the nearest integer, the value is approximately 55 J.