Question

When 18.5 J was added as heat to a particular ideal gas, the
volume of the gas changed from 36.8 cm^{3} to 59.4
cm^{3} 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}
m^{3}

pressure, p = 0.944 atm= 0.944 x 1.01 x 10^{5} Pa =
0.953 x 10^{5} Pa

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

= 18.5 - (0.953 x 10^{5} 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

P_{1}V_{1}=nRT_{1}

P_{2}V_{2}=nRT_{2}

delta T = T2 - T1 = (P/nR) (V_{2}-V_{1}) = 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

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