Question
Figure shows the variation of internal energy $$\left( U \right)$$ with the pressure $$\left( P \right)$$ of $$2.0\,mole$$ gas in cyclic process $$abcda.$$ The temperature of gas at $$c$$ and $$d$$ are $$300\,K$$ and $$500\,K.$$ Calculate the heat absorbed by the gas during the process.
Figure shows the variation of internal energy $$\left( U \right)$$ with the pressure $$\left( P \right)$$ of $$2.0\,mole$$ gas in cyclic process $$abcda.$$ The temperature of gas at $$c$$ and $$d$$ are $$300\,K$$ and $$500\,K.$$ Calculate the heat absorbed by the gas during the process.
A.
$$400\,R\ln 2$$
B.
$$200\,R\ln 2$$
C.
$$100\,R\ln 2$$
D.
$$300\,R\ln 2$$
Answer :
$$400\,R\ln 2$$
Solution :
Change in internal energy for cyclic process $$\left( {\Delta U} \right) = 0.$$
For process $$a \to b,\,\left( {P{\text{ - constant}}} \right)$$
$$\eqalign{ & {W_{a \to b}} = P\Delta V \cr & = nR\Delta T = - 400\,R \cr} $$
For process $$b \to c,\,\,\left( {T{\text{ - constant}}} \right)$$
$${W_{b \to c}} = - 2R\left( {300} \right)\ln 2$$
For process $$c \to d,\,\left( {P{\text{ - constant}}} \right)$$
$${W_{c \to d}} = + 400\,R$$
For process $$d \to a,\,\left( {T{\text{ - constant}}} \right)$$
$${W_{d \to a}} = + 2R\left( {500} \right)\ln 2$$
Net work $$\left( {\Delta W} \right) = {W_{a \to b}} + {W_{b \to c}} + {W_{c \to d}} + {W_{d \to a}}$$
$$\eqalign{ & \Delta W = 400\,R\ln 2 \cr & \therefore dQ = dU + dW,\,{\text{first law of thermodynamics}} \cr & \therefore dQ = 400\,R\ln 2. \cr} $$
Change in internal energy for cyclic process $$\left( {\Delta U} \right) = 0.$$
For process $$a \to b,\,\left( {P{\text{ - constant}}} \right)$$
$$\eqalign{ & {W_{a \to b}} = P\Delta V \cr & = nR\Delta T = - 400\,R \cr} $$
For process $$b \to c,\,\,\left( {T{\text{ - constant}}} \right)$$
$${W_{b \to c}} = - 2R\left( {300} \right)\ln 2$$
For process $$c \to d,\,\left( {P{\text{ - constant}}} \right)$$
$${W_{c \to d}} = + 400\,R$$
For process $$d \to a,\,\left( {T{\text{ - constant}}} \right)$$
$${W_{d \to a}} = + 2R\left( {500} \right)\ln 2$$
Net work $$\left( {\Delta W} \right) = {W_{a \to b}} + {W_{b \to c}} + {W_{c \to d}} + {W_{d \to a}}$$
$$\eqalign{ & \Delta W = 400\,R\ln 2 \cr & \therefore dQ = dU + dW,\,{\text{first law of thermodynamics}} \cr & \therefore dQ = 400\,R\ln 2. \cr} $$
