Question

Air at 1 atm and 20 0C occupies an initial volume of 1000 cm3 in a cylinder. The air is confined by a piston which has a constant restraining force so that the gas pressure always remains constant. Heat is added to the air until its temperature reaches 260 0C. Calculate (a) the heat added (b) the work done by the gas, (c) the change in internal energy of the gas.

Answer 1

Note : I have given answers in the order part c, part b and part a

let

P1 = 1 atm = 1.013*10^5 pa
T1 = 20 C

= 20 + 273

= 293 K
V1 = 1000 cm^3

= 1000*10^-6 m^3

P2 = 1 atm = 1.013*10^5 pa

T2 = 260 C

= 260 + 273

= 533 K

let V2 is the final colume.

at constant pressure, V/T = constant

V2/T2 = V1/T1

V2 = V1*(T2/T1)

= 1000*10^-6*(533/293)

= 1.819*10^-3 m^3

part C) The change in internal energy of the gas, delta_U = (3/2)*n*R*(T2 - T1)

= (3/2)*P2*V2 - (3/2)*P1*V1

= (3/2)*P*(V2 - V1)

= (3/2)*1.013*10^5*(1.819*10^-3 - 1000*10^-6)

= 124 J <<<<<<<------------Answer for part c)

part b) Work done by the gas, W = P1*(V2 - V1)

= 1.013*10^5*(1.819*10^-3 - 1000*10^-6)

= 83 J <<<<<<<------------Answer for part b)

part a) Using first law of thermodynamics,

Heat added,

dQ = W + dU

= 124 + 83

= 207 J <<<<<<<------------Answer for part a)