Question

The ratio of the molar heat capacities of an ideal gas is Cp/Cv = 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

The change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K by keeping the pressure and volume constant and also adiabatically is  2490 J.

Explanation:

Step 1:

$\left(\frac{c_{p}}{c_{v}}\right)=76$

Amount of Gas Moles,n = 1 mol

Temperature change for the gas, ∆T = 50 K

Step 2:

(a) Pressure sustaining constant: Using Thermodynamics First Law,

$\begin{array}{l}{d Q=d U+d W} \\{\Delta T=50 K \text { and } \gamma=\frac{7}{6}} \\{d Q=d U+d W}\end{array}$

Work done, dW = PdV

Work is done as pressure is maintained constant = P(∆V)

Use of perfect equation for gas  PV = nRT,

P(ΔV) = nR(∆T)

dW = nR(∆T)

In case of constant pressure,  

$d Q=n c_{p} d T$

Step 3:

These values are replaced by the thermodynamics first law,

$n c_{p} d T=d U+R d T |$

$d U=n c_{p} d T-R d T$

Using  $\frac{c_{p}}{c_{v}}=\gamma$ and

$c_{p}-c_{v}=R, \text { we get }$

$d U=1 \times \frac{R y}{\gamma-1} \times d T-R d T$

= 7 RdT - RdT

= 7 RdT - RdT = 6 RdT

= 6 × 8.3 × 50 = 2490 J

Step 4:

(b) Constant volume-keeping:

Work done = 0

Use of the first thermodynamics law,

dU  = dQ

$\begin{array}{l}{d U=n C_{v} d T} \\{=1 \times \frac{R}{\gamma-1} \times d T=1 \times\left(\frac{8.3}{\frac{7}{6}-1}\right) \times 50}\end{array}$

= 8.3 × 50 × 6 = 2490 J

Step 5:

(c) Adiabatically, dQ = 0

Use of the first thermodynamics law,

dU = - dW

$=\left[\frac{n \times R}{\gamma-1}\left(T_{1}-T_{2}\right)\right]$

$=\frac{1 \times 8.3}{\frac{7}{6}-1}=\left(T_{1}-T_{2}\right)$

= 8.3 × 6 × 50 = 2490 J