Question

In this problem, 0.90 mole of a monatomic ideal gas is initially at 285 K and 1 atm.

(a) What is its initial internal energy?

_____ kJ

(b) Find its final internal energy and the work done by the gas when 420 J of heat are added at constant pressure.

final internal energy ________kJ

work done by the gas _______kJ

(c) Find the same quantities when 420 J of heat are added at constant volume.

finale internal energy ________kJ

work done by the gas ________kJ

Answer 1

Internal enrgy of gas U = f/2 ( nRT)
f is degree of freedom, n is number of moles and T is tempreture of gas in Kelvin, R=8.3, is gas constant
f = 3 for monoatomic gas
a) So U_i = 3x0.9x8.3x285/2= 3.2 KJ
b) Heat given to gas at constant pressure is
Q = n C_p ( dleta T) C_p is molar specific heat at constant pressure and = (f+2)R/2 , (delta T) is chsnge in tempreture during process
hence 420 = 0.9 x 5 x 8.3x (delta T)/2
Hence (delta T) = 22.5 K
Work done by gas at constant pressure is
W = n R (delta T) = 0.9x 8.3 x22.5 = 0.17 KJ
Q = (delta U) + W
hence (delta U) = 0.42 - 0.17 = 0.25 KJ
So U_f = 3.2 + 0.25 = 3.45 KJ
c) Work done by gas during constant volume process = 0
Hence at constant volume (delta U) = Q = 0.42 KJ
and U_f = 3.62 KJ