Two moles of an ideal gas are changed from their initial state(16 atm, 6L) to final state(4 atm, 15L) in such a way that this change can be represented by a straight line in P-V curve. Let 'y' atm-L be the maximum average translational kinetic energy possessed by the gas sample during the above change. [Take : R = \frac{1}{12} atm-L-mol–1K

Engineering

Chemistry

Kinetic Theory of Gases KTG

Question

Two moles of an ideal gas are changed from their initial state(16 atm, 6L) to final state(4 atm, 15L) in such a way that this change can be represented by a straight line in P-V curve. Let 'y' atm-L be the maximum average translational kinetic energy possessed by the gas sample during the above change.

[Take : R = $\frac{1}{12}$ atm-L-mol^–1K^–1]

T_max = 648K

${KE}_{\max} = \frac{3}{2} {nRT}_{\max} = \frac{3}{2} \times 2 \times \frac{1}{12} \times 648 = 162 atm - L = \frac{162}{81} = 2$

Understanding the Problem

We are given an ideal gas undergoing a linear change on a P-V diagram from initial state (16 atm, 6 L) to final state (4 atm, 15 L). We need to find the maximum average translational kinetic energy (denoted as 'y' atm-L) during this process. The average translational kinetic energy per mole for an ideal gas is given by $\frac{3}{2} R T$ . Since there are 2 moles, the total kinetic energy is $K E = 2 \times \frac{3}{2} R T = 3 R T$ . Given R = $\frac{1}{12}$ atm-L/mol-K, we have KE = $3 \times \frac{1}{12} T = \frac{T}{4}$ atm-L. So, KE is directly proportional to temperature T. To maximize KE, we need to maximize T during the process.

Step-by-Step Solution

Step 1: Find the Equation of the Straight Line on P-V Diagram

The process is a straight line from (V_i=6 L, P_i=16 atm) to (V_f=15 L, P_f=4 atm). The slope (m) is:

m = \frac{ΔP}{ΔV} = \frac{4 - 16}{15 - 6} = \frac{- 12}{9} = - \frac{4}{3}

Using point-slope form with point (6,16):

P - 16 = - \frac{4}{3} (V - 6)

Simplifying to get P as a function of V:

P = - \frac{4}{3} V + 8 + 16 = - \frac{4}{3} V + 24

So, the equation is:

P = 24 - \frac{4}{3} V

Step 2: Express Temperature T as a Function of Volume V

For an ideal gas, PV = nRT. With n=2 and R= $\frac{1}{12}$ , we have:

P V = 2 \times \frac{1}{12} T = \frac{T}{6}

Therefore, T = 6PV.

Substitute P from the line equation:

T = 6 V (24 - \frac{4}{3} V) = 144 V - 8 V^{2}

So, T(V) = -8V² + 144V.

Step 3: Find the Volume V where Temperature is Maximum

The expression T(V) = -8V² + 144V is a quadratic equation opening downwards (a = -8 < 0), so its maximum occurs at the vertex. The vertex V-coordinate is:

V = - \frac{b}{2 a} = - \frac{144}{2 \times (- 8)} = - \frac{144}{- 16} = 9

The maximum temperature occurs at V = 9 L.

Step 4: Find the Pressure and Temperature at V=9 L

Substitute V=9 into the line equation to find P:

P = 24 - \frac{4}{3} \times 9 = 24 - 12 = 12

P = 12 atm.

Now calculate T using T = 6PV:

T = 6 \times 12 \times 9 = 648

T = 648 K.

Step 5: Calculate the Maximum Kinetic Energy

As derived earlier, the total translational kinetic energy for 2 moles is KE = T/4 atm-L.

K E = \frac{648}{4} = 162

Therefore, the maximum average translational kinetic energy y = 162 atm-L.

Final Answer: y = 162

Key Formulae and Theory

Ideal Gas Law: $P V = n R T$

Average Translational Kinetic Energy per mole: $\bar{KE} = \frac{3}{2} R T$

For 'n' moles: $KE = n \times \frac{3}{2} R T$

Vertex of a Parabola: For a quadratic function f(x)=ax²+bx+c, the x-coordinate of the vertex is $x = - \frac{b}{2 a}$ . If a < 0, this point is a maximum.

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Related Topics

Key Formulae and Theory

Detailed Solution

Understanding the Problem

Step-by-Step Solution

Related Topics

Key Formulae and Theory

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