Name: Sign Conventions and Work Done in Thermodynamics
Uploaded: 2019-07-19
Duration: 6 min 11 s
Description: In the field of thermodynamics, either heat is given to a system or by the system. Similarly, work may be done by the system or on the system. Let's learn more!

Question

For one mole of an ideal gas at $300 K$ occupies a volume of $0.36 m^{3}$ at a pressure of $2 atm$ . The gas expands adiabatically until its volume becomes $1.44 m^{3}$ . Next, the gas is compressed isobarically to its original volume. Finally, the pressure is increased isochorically until the gas returns to as initial state (see figure).

(a) Determine the temperature at the point $B and C$ .

(b) The work done during the process $A to B$ .

(Assume, $γ = 1.5, R = 8 J {mol}^{- 1} K^{- 1}$ .)

Accepted Answer

From the given graph, we need to solve the different questions asked.

For part (a) of the question, let us examine the process $A B$ from the graph. We can identify that the process is adiabatic.

So,

$T V^{γ - 1} = C$

$T_{A} {V_{A}}^{γ - 1} = T_{B} {V_{B}}^{γ - 1}$

$\frac{T_{B}}{T_{A}} = {(\frac{V_{A}}{V_{B}})}^{γ - 1} = {(\frac{0.36}{1.44})}^{1.5 - 1} {(\frac{36}{144})}^{(\frac{1}{2})} = \frac{1}{2} \frac{T_{B}}{T_{A}} = \frac{1}{2}$

So, from this information, we can take out $T_{B}$ ,

$T_{B} = \frac{T_{A}}{2} \Rightarrow T_{B} = \frac{300}{2} = 150 K$

Similarly, for the process $B C$ , it is an isochoric process,

$V \propto T \frac{V_{B}}{V_{C}} = \frac{T_{B}}{T_{C}}$

$T_{C} = T_{B} \frac{V_{C}}{V_{B}} = 150 \times \frac{36}{144} \Rightarrow T_{C} = \frac{150}{4} = 37.5 K$

(b) Solving for second part of the question,

${(W D)}_{AB} = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = \frac{n R T_{1} - n R T_{2}}{γ - 1}$

$= \frac{n R (T_{A} - T_{B})}{γ - 1} = \frac{1 \times 8 \times (300 - 150)}{1.5 - 1} = \frac{8 \times 150}{0.5} = 8 \times 150 \times 2 = 2400 J$

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