Answer to Question #190720 in Molecular Physics | Thermodynamics for Ahmed

Question

Suppose you are given a one mole of ideal gas, in an initial state p=6 atm, v=3L, and in a clock wise direction, you maneuvered the system reverisbly in a path given by (v-6)²

+(p-6)² =16 in the pv plane

Deduce the amount of work done by the gas as a result of this process

Find out the change if the internal energy of the gas

Calculate the minimum and maximum temperature attained by the gas during the cycle

Answer 1

$p = 6\ atm\\ V = 3\ L$

$path = (V-6)^2 +(p-6)^2 =16$

maximum value of V when p = 6 atm

$(V-6)² + (6-6)² = 16\\ (V-6)² = 16\\ V-6 = 4\\ V= 10\ L$

maximum value of p when V = 3 L

$(3-6)²+ (p -6)² = 16\\ (-3)² +(p-6)² =16\\ 9+ (p-6)²= 16\\ (p-6)² = 7\\ p-6 =\sqrt{7}\\ p= 6 +\sqrt7= 8.65 \ atm$

Work done = $\dfrac12(V_f-V_i)(p_f-p_i)$

$=\dfrac12(10-3)(8.65-6) = 9.275\ J$

$∆U = Q-W.D = 0-9.275J$

$\therefore ∆U = -9.275\ J$