One mole of monoatomic gas is taken through cyclic process shown below. $T_{A} = 300 K .$ Process $A B$ is defined as PT = constant.
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Change in internal energy in process $C A$ is $- 900 R .$

Heat transferred in the process $B C$ is $2000 R .$

Work done in process $A B$ is $- 400 R .$

Change in internal energy in process $C A$ is $900 R .$

answer is A, C, D.

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Detailed Solution

Process AB :
$P T =$ constant
$n R T^{2} =$ constant
$W = \int_{A}^{B} P d V = \int_{A}^{B} \frac{c o n s t a n t}{T} d V \Rightarrow \frac{d V}{d T} = \frac{2 n R T}{c o n s t a n t} W = \int_{300}^{100} \frac{c o n s t a n t}{T} . \frac{2 n R T}{c o n s t a n t} d t \Rightarrow \frac{P_{A}}{P_{B}} = \frac{T_{B}}{T_{A}} \Rightarrow \frac{1}{3} = \frac{T_{B}}{T_{A}} \Rightarrow T_{B} = \frac{300}{3} = 100 \Rightarrow W = 2 n R (100 - 300) \Rightarrow W_{A B} = - 400 n R$
Process CA :
Isochoric $\frac{P}{T}$ constant :
$\frac{T_{A}}{T_{C}} = \frac{P_{A}}{P_{C}} \frac{T_{A}}{T_{C}} = \frac{P_{A}}{P_{B}} \Rightarrow \frac{T_{A}}{T_{C}} = \frac{1}{3} \Rightarrow T_{C} = 3 T_{A} T_{C} = 900 R \Rightarrow Δ U = n C_{V} Δ T = (1) \frac{3}{2} R \times (T_{A} - T_{C}) = \frac{3}{2} R \times (300 - 900) = \frac{3}{2} R \times - 600 = - 900 R \Rightarrow | Δ U | = (900 R)$
Process BC :
Isobaric
$Q = n C_{P} Δ T \Rightarrow Q = (1) \frac{5}{2} R \times (T_{C} - T_{B}) Q = \frac{5}{2} R \times (900 - 100) \Rightarrow Q = \frac{5}{2} R \times 800 \Rightarrow Q = 2000 R$