Question

Consider a reversible thermodynamic process involving one mole of ideal gas represented by a line on the $P-V$ diagram connecting two states X (4 bar, 1 L, $\text{T}_1$) and Y (1 bar, 2 L, $\text{T}_2$) as shown in the figure:

During the process the change in temperature (in K) as a function of volume (in L) is best represented as:

• A. A parabolic curve opening downwards with a maximum temperature.
• B. A hyperbolic curve decreasing monotonically.
• C. A straight line with a negative slope.
• D. A parabolic curve opening upwards with a minimum temperature.

Hint:

Any straight-line process on a $P-V$ diagram with a negative slope will always yield a quadratic relationship for temperature as a function of volume ($T \propto -V^2$), which manifests as a downward-opening parabola.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the mathematical representation of the temperature (T) as a function of volume (V) for an ideal gas undergoing a linear transition from state X ($1\ \text{L}$, $4\ \text{bar}$) to state Y ($2\ \text{L}$, $1\ \text{bar}$).

Step 2: Key Formula or Approach:
1. The equation of a straight line in the $P-V$ plane is:
\[ P = mV + c \]
2. The ideal gas equation for 1 mole of gas is:
\[ PV = RT \implies T = \frac{PV}{R} \]

Step 3: Detailed Explanation:

• Let us first find the equation of the linear path connecting state X $(V_1, P_1) = (1, 4)$ and state Y $(V_2, P_2) = (2, 1)$ on the $P-V$ diagram.

• The slope ($m$) of this line is:
\[ m = \frac{P_2 - P_1}{V_2 - V_1} = \frac{1 - 4}{2 - 1} = -3 \]

• Using the point-slope form with state Y $(2, 1)$:
\[ P - 1 = -3(V - 2) \implies P = -3V + 7 \]

• Now, let us substitute this expression for pressure into the ideal gas equation:
\[ T = \frac{PV}{R} = \frac{(-3V + 7)V}{R} = \frac{-3V^2 + 7V}{R} \]

• This equation defines the temperature as a quadratic function of volume ($\text{T} \propto -V^2$).

• Mathematically, this is a parabola that opens downwards because the coefficient of the $V^2$ term is negative.

• Let us find the volume at which the maximum temperature occurs:
\[ \frac{dT}{dV} = \frac{-6V + 7}{R} = 0 \implies V = \frac{7}{6} \approx 1.17\ \text{L} \]

• Since $1.17\ \text{L}$ lies inside our experimental range ($1\ \text{L}$ to $2\ \text{L}$), the temperature must first increase to a peak at $1.17\ \text{L}$ and then decrease as we continue to $2\ \text{L}$.


Step 4: Final Answer:
Therefore, the temperature as a function of volume is best represented by the downward-opening parabola shown in Option A.