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

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