Question

Suppose 1300 J of heat are added to 4.3 mol of argon gas at a constant pressure of 120 kPa. (Assume that the argon can be treated as an ideal monatomic gas.) (a) Find the change in internal energy. J (b) Find the change in temperature for this gas. K (c) Calculate the change in volume of the gas.

Answer 1

Added heat = 1300 J
Number of moles of argon, n = 4.3
Pressure P = 120 kPa = 120 * 10 ^ 3 Pa

we know Q = n*Cp*dT ------ ( 1)
where Cp= specific heat at constant pressure = ( 5/ 2) R
R = gas constant = 8.314 J / mol K
Cp = 20.785 J / mol K

from eq ( 1) , dT = Q / n*Cp
=> dt = 1300/(5/2*8.314*4.3)
= 14.54 K

(a). Change in internal energy dU = n*Cv*dT
where Cv = specific heat at constant volume = ( 3/ 2) R
dU = n*Cv*dT
= 4.3*(3/2*8.314)*14.54
= 780 J

(b). Change in temperature of the gas dT = 14.54 K

(c).Change in volume of ht egas dV = ?
we know at constant pressure P *dV = n*R*dT
from this dV = nR dT / P
= 4.3*8.314*14.54/(120 * 10 ^ 3)
= 4.33*10^-3 m^ 3