A volume of air (assumed to be an ideal gas) is first cooled without changing it

Question

A volume of air (assumed to be an ideal gas) is first cooled without changing its volume and then expanded without changing its pressure, as shown by the path abc, a) How does the final temperature of the gas compare with its initial temperature? b)How much heat does the air exchange with its surrounding during the process abc? Does the air absorb heat or release heat during this process? Explain c)If the air instead expands from state a to state c by the straight line path shown how much heat does it exchange with its surroundings? values of p in Pa for ab is from 3.0x10^5 to 1.0x10^5 values of V in m^3 from b to c is from .02 to .06

Explanation / Answer

1) let the temperature at a be T.. now a--->b is isochoric process and pressure is becoming 1/3 times of that at a apply, P/T=constant we get temperature at b as T/3 and, b----->c is isobaric process and volume is becoming three times to that at b apply, V/T=constant we get temperature at c = T/3*3 = T hence temperature at c equals to that at a. 2)the internal energy of a gas is a function of temperature only. As temperature at a and c are same their internal energy at a and c is also same. 3) i) according to the first law of thermodynamics, Q = U + W ........ (1) where Q = heat absorbed by the system ( to be calculated) U = change in internal energy of system ( here it is zero as there is no change in internal energy between points a and c) W = work done by the system (area under the p-v diagram) now W = W(for a--->b)+W(for a---->c) ........... (2) now, W(for a----->b) = 0 (W = pdv and for isochoric process dv=0 so W=0) W(for b----->c) = pdv = p_b(v_b-v_c) = 2*10^5(0.06-0.02) = 8*10^3 J = 8 kJ (for isobaric process pressure is constant so we use such a formula) now, putting these values in (2) we get, W = 0 + 8 kJ = 8kJ now we put values of U and W in (1) we get, Q = 0 + 8 kJ = 8 kJ this is the amount of heat exchanged. Also the sign of Q is positive, it means that heat is being absorbed by the system. 4)again we apply the same rule and U = 0, but W equals area under the line ac.. so, W = 1/2(4*10^5*0.04) + 2*10^5(0.06-0.02) = 16 kJ so Q = 16 kJ ( absorbed by the system)

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