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  • Consider the P − V diagram shown in Figure . The diagram represents the process of n moles of an ideal gas undergoing a thermodynamic cycle between points A and B. Your task is to find the ma

Consider the P − V diagram shown in Figure . The diagram represents the process of n moles of an ideal gas undergoing a thermodynamic cycle between points A and B. Your task is to find the maximum temperature of the gas during this cycle.

Option: 1

125k


Option: 2

170 k


Option: 3

200 k


Option: 4

20 k


Answers (1)

best_answer

To find the maximum temperature of the gas between points A and B, we need to determine the path of the process that corresponds to the maximum temperature. The temperature of an ideal gas is related to its pressure and volume by the ideal gas law:

P V = nRT,        (1)

where P is the pressure, V is the volume, n is the number of moles of the gas, R is the ideal gas constant, and T is the temperature. Looking at the P − V diagram, we observe that the maximum temperature occurs at the point where the gas is compressed the most while maintaining constant pressure. This corresponds to the steepest downward slope in the diagram.

Using this information, we can see that the segment from A to B with the steepest slope is the vertical segment on the right side. Along this segment, the volume decreases while the pressure remains constant. Therefore, we can write:

?V = VB − VA,         (2)

where VB is the volume at point B and VA is the volume at point A. Since the process is isobaric (constant pressure), we can write:

\frac{\Delta V}{V}=\frac{\Delta T}{T}     (3)

where T is the temperature.

Solving for ?T, we get:

\Delta T=T_B-T_A=\frac{\Delta V}{V} \cdot T,     (4)

where TB is the temperature at point B and TA is the temperature at point A. Now we can plug in the values from the diagram. Let’s assume the initial volume at point A is VA and the final volume at point B is VB. Given that the volume decreases from A to B, ?V = VB − VA. Substituting into the equation:

\Delta T=\frac{V_B-V_A}{V_A} \cdot T .        (5)

The maximum temperature Tmax occurs when ?T is maximum. To achieve this, VB should be as small as possible while VA should be as large as possible. Since the vertical segment represents the steepest slope, it implies that VB is the smallest and VA is the largest along this segment. Therefore, the maximum temperature Tmax is given by:

T_{\max }=\frac{V_{\min }-V_{\max }}{V_{\max }} \cdot T,           (6)

where Vmin is the smallest volume and Vmax is the largest volume along the vertical segment. Now, using the values from the diagram, we can calculate the maximum temperature:

T_{\max }=\frac{V_B-V_A}{V_A} \cdot T .

Substituting the numerical values and solving for Tmax, we get:

T_{\max }=\frac{1.0 \mathrm{~L}-0.4 \mathrm{~L}}{0.4 \mathrm{~L}} \cdot 300 \mathrm{~K}=125 \mathrm{~K}          (8)

Therefore, the maximum temperature of the gas between points A and B is 125 K. Therefore, the correct option is 1.

 

 

Posted by

manish painkra

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