First slide

Thermodynamic processes

Question

One mole of monoatomic gas is taken through cyclic process shown below. T_A = 300 K. Process AB is defined as PT = constant.

Moderate

A

Work done in process AB is - 400 R.
B

Change in internal energy in process CA is 900 R
C

Heat transferred in the process BC is 2000 R.
D

Change in internal energy in process CA is - 900 R.

Solution

(a) Process AB :
   $\begin{array}{l} PT = constant \\ \Rightarrow \frac{{nRT}^{2}}{V} = constant \\ W = \int_{A}^{B} PdV = \int_{A}^{B} \frac{Constant}{T} dV \\ \Rightarrow \frac{dV}{dT} = \frac{2 nRT}{Constant} \\ W = \int_{300}^{100} \frac{Constant}{T} \cdot \frac{2 nRT}{Constant} dt \\ \Rightarrow \frac{P_{A}}{P_{B}} = \frac{T_{B}}{T_{A}} \\ \Rightarrow \frac{1}{3} = \frac{T_{B}}{T_{A}} \\ \Rightarrow T_{B} = \frac{300}{3} = 100 \\ \Rightarrow W = 2 nR (100 - 300) \\ \Rightarrow W_{AB} = - 400 nR \end{array}$
(b) Process CA :
$Isochoric, so \frac{P}{T} is constant :$
   $\begin{array}{l} \frac{T_{A}}{T_{C}} = \frac{P_{A}}{P_{C}} \\ \frac{T_{A}}{T_{C}} = \frac{P_{A}}{P_{B}} \\ \Rightarrow \frac{T_{A}}{T_{C}} = \frac{1}{3} \\ \Rightarrow T_{C} = 3 T_{A} \\ T_{C} = 900 R \\ \Rightarrow ΔU = {nC}_{V} ΔT \\ = (1) \frac{3}{2} R \times (T_{A} - T_{C}) \\ = \frac{3}{2} R \times (300 - 900) \\ = \frac{3}{2} R \times - 600 = - 900 R \\ \Rightarrow | ΔU | = (900 R) \end{array}$
(c) Process BC:
   $Isobaric$
   $\begin{array}{l} Q = {nC}_{P} ΔT \\ \Rightarrow Q = (1) \frac{5}{2} R \times (T_{C} - T_{B}) \\ Q = \frac{5}{2} R \times (900 - 100) \\ \Rightarrow Q = \frac{5}{2} R \times 800 \\ \Rightarrow Q = 2000 R \end{array}$

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