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

One mole of a diatomic ideal gas $(γ = 1.4)$ is taken through a cyclic process starting from point $A$ . The process $A \to B$ is an adiabatic compression. $B \to C$ is isobaric expansion. $C \to D$ an adiabatic expansion and $D \to A$ is isochoric as shown in P-V diagram. The volume ratios are $\frac{V_{A}}{V_{B}} = 16 & \frac{V_{C}}{V_{B}} = 2$ and the temperature at $A$ is $T_{A} = 300 K$ . Calculate the temperature of the gas at the points $B$ and $D$ and find the efficiency of the cycle.

Accepted Answer

It is given that $A \to B$ is an adiabatic process. Hence, we can use the expression for adiabatic process as given below:

$T_{A} {(V_{A})}^{(γ - 1)} = T_{B} {(V_{B})}^{(γ - 1)} \Rightarrow T_{B} = T_{A} \times {(\frac{V_{A}}{V_{B}})}^{(γ - 1)} \Rightarrow T_{B} = 300 \times {(16)}^{(1.4 - 1)} \approx 909 K$

Similarly, $B$ is also an adiabatic process, we can use the equation for adiabatic process for this change again:

$T_{C} {(V_{C})}^{(γ - 1)} = T_{D} {(V_{D})}^{(γ - 1)} \Rightarrow T_{D} = T_{C} \times {(\frac{V_{C}}{V_{D}})}^{(γ - 1)} . . . (1)$

Now, we need to calculate $T_{C}$ and $\frac{V_{C}}{V_{D}}$ . We know that $B \to C$ is an isobaric process $\frac{V_{C}}{T_{C}} = \frac{V_{B}}{T_{B}} \Rightarrow T_{C} = T_{B} \times \frac{V_{C}}{V_{B}} \Rightarrow T_{C} = 909 \times 2 = 1818 K$

$\frac{V_{A}}{V_{B}} = 16, \frac{V_{C}}{V_{B}} = 2 \Rightarrow \frac{V_{A}}{V_{C}} = 8, \Rightarrow \frac{V_{D}}{V_{C}} = 8 (∵ V_{A} = V_{D})$

From equation $(1)$ , $T_{D} = 1818 \times {(\frac{1}{8})}^{(1.4 - 1)} \Rightarrow T_{D} = 791 K$

Work done in process $A B$ $W_{A B} = \frac{n R}{γ - 1} (T_{A} - T_{B}) = \frac{1 \times 8.3}{1.4 - 1} (300 - 909) \Rightarrow W_{A B} = - 12637 J$

Work done in process $B C$ $W_{B C} = n C_{p} ∆ T - n C_{v} ∆ T \Rightarrow 1 \times 8.3 \times (1818 - 909) \Rightarrow W_{B C} = 7545 J$

Work done in process $C D$ $W_{C D} = \frac{n R}{γ - 1} (T_{c} - T_{D}) = \frac{1 \times 8.3}{1.4 - 1} (1818 - 791) \Rightarrow W_{C D} = 21310 J$

Work done in process $D A$ is zero, no change in volume. $W_{D A} = 0$

The net work done in all process $W_{n e t} = W_{A B} + W_{B C} + W_{C D} + W_{D A} = 16218 J$

Heat is supplied in process $B C$ $H_{B C} = n C_{p} ∆ T = 1 \times \frac{7 R}{2} (1818 - 909) \Rightarrow H_{B C} = 26232 J$

The efficiency of the cycle $η = \frac{W}{Q} \Rightarrow η = \frac{16218}{26232} \Rightarrow η = 0.62$

Important Questions on Thermodynamics

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