Question

The ratio of the molar heat capacities of an ideal gas is C_p/C_v = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K (a) keeping the pressure constant (b) keeping the volume constant and (c) adiabatically.

Answer 1

Given:
$\left(\frac{{\mathrm{C}}_p}{{\mathrm{C}}_v}\right)=\frac{7}{6}$
Number of moles of the gas, n = 1 mol
Change in temperature of the gas, ∆T = 50 K

(a) Keeping the pressure constant:
Using the first law of thermodynamics,
dQ = dU + dW
∆T = 50 K and $\gamma =\frac{7}{6}$
dQ = dU + dW
Work done, dW = PdV
As pressure is kept constant, work done = P($\mathit{∆}V$)
Using the ideal gas equation PV = nRT,
P($\mathit{∆}V$) = nR($∆$T)
⇒ dW = nR($∆$T)
At constant pressure, dQ = nC_p dT
Substituting these values in the first law of thermodynamics, we get
nC_p dT = dU + RdT
⇒ dU = nC_p dT − RdT
Using $\frac{C_p}{C_V}\mathit{=}\gamma \mathit{}\mathit{}\mathit{}\mathit{}\mathit{\text{and}}\mathit{}{C}_P\mathit{-}{C}_V\mathit{=}R,\text{we get}$
$dU=1\times \frac{R\gamma }{(\gamma -1)}\times d\mathrm{T}-Rd\mathrm{T}$
= 7 RdT − RdT
= 7 RdT − RdT = 6 RdT
= 6 × 8.3 × 50 = 2490 J

(b) Keeping volume constant:
Work done = 0
Using the first law of thermodynamics,
dU = dQ
dU = nC_v dT
$$\begin{gathered} =1\times \frac{R}{\gamma -1}\times d\mathrm{T} \\ =1\times \left(\frac{8.3}{{\displaystyle \frac{7}{6}}-1}\right)\times 50 \end{gathered}$$
= 8.3 × 50 × 6 = 2490 J

(c) Adiabatically, dQ = 0
Using the first law of thermodynamics, we get
dU = − dW
$$\begin{gathered} =\left[\frac{n\times R}{\gamma -1}(T_1-T_2)\right] \\ =\frac{1\times 8.3}{7/6-1}=(T_2-T_1) \end{gathered}$$
= 8.3 × 6 × 50 = 2490 J