Question

Two moles of an ideal gas is changed from its initial state (16 atm, 6L) to final state (4 atm, 15L) in sucha way that this change can be represented by a straight line in P–V curve. The maximum temperature attained by the gas during the above change is : (Take R =1/12L atm K–1 mol–1)
(A) 324 K (B) 648 K (C) 1296 K (D) 972 K

Answer 1

Question:

Two moles of an ideal gas is changed from its initial state (16 atm, 6L) to final state (4 atm, 15L) in sucha way that this change can be represented by a straight line in P–V curve. The maximum temperature attained by the gas during the above change is : (Take R =1/12L atm K–1 mol–1)

(A) 324 K (B) 648 K (C) 1296 K (D) 972 K

Solution:

Equation of straight line is given as:

$y-y_1 = \dfrac{y_2-y_1}{x_2-x_1} (x-x_1)$

here, y = P,x=V, , y₁ = 16 , y₂ = 4 , x₁ = 6 , x₂ = 16 , then,

$P - 16 = \dfrac{4-16}{15-6}(V-6)\\\\3P+4V = 72$

We know, $T_{max} = \dfrac{(PV)_{max}}{nR}$

For $(PV)_{max}, 3P = \dfrac{72}{2}\text{ and }4V = \dfrac{72}{2}$

or, P = 12 , V = 9

Hence, $T_{max} = \dfrac{12\times9}{1\times1/12} = 648 \ K$

Therefore, option B is correct.