In the given (V-T) diagram, what is the relation between pressures $P_1$ and $P_2$ ?
- $P_2 = P_1$
- $P_2 > P_1$
- $P_2 < P_1$
- Cannot be predicted
The Correct Option is C
Solution and Explanation
To determine the relation between the pressures \(P_1\) and \(P_2\) based on a Volume-Temperature (V-T) diagram, we can use the principles of the ideal gas law, which is given by:
\(PV = nRT\)
Where:
- \(P\) = Pressure
- \(V\) = Volume
- \(T\) = Temperature (in Kelvin)
- \(n\) = Number of moles
- \(R\) = Universal gas constant
For a given mass of an ideal gas, the number of moles \(n\) and the gas constant \(R\) are constant. Therefore, the ideal gas law can be rearranged for pressure as:
\(P = \frac{nRT}{V}\)
From the ideal gas law, we know at two states:
- State 1: \(P_1V_1 = nRT_1\)
- State 2: \(P_2V_2 = nRT_2\)
The relationship between pressures can be derived if there is a relation between the volumes and temperatures. From the options provided, let's analyze to identify conditions:
If the Volume-Temperature graph shows a directly proportional relationship (for example, constant pressure process), and assume volumes vary directly with temperatures, we have:
\(\frac{V_1}{T_1} = \frac{V_2}{T_2}\)
This implies an isobaric process where \(P_1 = P_2\). However, if not directly proportional, the pressures could be different.
Without any other information on how \(V\) is varying with respect to \(T\) or specific graphical analysis showing a constant or directly proportional sequence, pressure comparison specifically \(P_2 < P_1\) implies that volumes may be the same while temperature is higher in state 1 than state 2, assuming consistent conditions.
Therefore, if the V-T graph implies lesser volume at higher temperatures, \(P_2\) is less than \(P_1\), corresponding with the correct answer:
\(P_2 < P_1\)
