Question

When 18.5 J was added as heat to a particular ideal gas, the volume of the gas changed from 36.8 cm³ to 59.4 cm³ while the pressure remained constant at 0.944 atm. (a) By how much did the internal energy of the gas change? If the quantity of gas present is 1.38 x 10^-3 mol, find the molar specific heat of the gas at (b) constant pressure and (c) constant volume.

Answer 1

According ot first law of thermodynamics

We take heat added to the system as positive quantity and work done by the system as positive quantity,

dQ= 18.5 J

dV = 59.4-36.8 = 22.6 cm3 = 22.6 x 10^-6 m³

pressure, p = 0.944 atm= 0.944 x 1.01 x 10⁵ Pa = 0.953 x 10⁵ Pa

(a) Change in internal energy, dU = dQ -pdV

= 18.5 - (0.953 x 10⁵ x 22.6 x 10^-6) = 18.5 - 2.15 = 16.35 J

(b), (c) We know that the specific heat capacity at constant pressure is given as

Q = nC_p deltaT

Since it is an ideal gas

P₁V₁=nRT₁

P₂V₂=nRT₂

delta T = T2 - T1 = (P/nR) (V₂-V₁) = 187.8 K

C_p = (Q/n delta T)

C_p = C_v +R

C_v = C_p -R

C_v = 71.3-8.314 = 62.98 J/mol K