A sample of CO(g) whose mass is 7.31 g is heated from 25.0°C to 197°C at a constant pressure of 4.62 bar.

Question

A sample of CO(g) whose mass is 7.31 g is heated from 25.0°C to 197°C at a constant pressure of 4.62 bar. Calculate w (in J) for this process if q = 1308.1 J.

Answer 1

Let us find the change in volume.

PV = nRT

V = nRT/P

n = moles = 7.31 g CO x 1 mol CO / 28 g = 0.261 moles

R = gas constant = 0.0821 Latm/Kmol

T = temperature in K = 25 + 273 = 298K

P = pressure in atm = 4.62 bar x ~1 atm / 1 bar = 4.62 atm

V = volume in liters

V = (0.261)(0.0821)(298) / 4.62

V = 1.38 L

work (w) when pressure is constant = -P∆V or the product of the pressure times the change in volume.

We know the initial volume, so we now must find the change in volume (∆V)

V1/T1 = V2/T2

1.38 L / 298 K = V2 / 470 K

V2 = 2.18 L

∆V = 2.18 L - 1.38 L = 0.80 L

w = -P∆V

w = -(4.62 atm)(0.80 L) = -3.70 Latm

converting this to joules, we have ...

w = -3.70 Latm x 101.325 J / Latm = -375 J

The sign is negative because the system has done work on the surrounding by an expansion.

However, q is positive 1308 J meaning heat was added to the system (endothermic).

Thus ...

∆E = q + w

∆E= 1308 J - 375 J = +933 J