Question
A thermodynamic process is shown in the figure. The pressure and volumes corresponding to some points in the figure are
A thermodynamic process is shown in the figure. The pressure and volumes corresponding to some points in the figure are
$$\eqalign{
& {p_A} = 3 \times {10^4}pa,\,\,{V_A} = 2 \times {10^{ - 3}}{m^3} \cr
& {p_B} = 8 \times {10^4}pa,\,\,{V_B} = 5 \times {10^{ - 3}}{m^3} \cr} $$
In process $$AB,600\,J$$ of heat is added to the system and in process $$BC,200\,J$$ of heat is added to the system. The change in internal energy of the system in process $$AC$$ would be
A.
$$560\,J$$
B.
$$800\,J$$
C.
$$600\,J$$
D.
$$640\,J$$
Answer :
$$560\,J$$
Solution :
For path $$AB,$$ applying first law of thermodynamics
$$\eqalign{ & dQ = dU + dW\,\,{\text{or}}\,\,dQ = {U_B} - {U_A} + dW \cr & {\text{or}}\,\,600 = {U_B} - {U_A} + 0\,\,\left( {{\text{for}}\,{\text{isochoric}}\,{\text{process}}\,dV = 0\,\,{\text{so,}}\,\,dW = 0} \right) \cr & {U_B} - {U_A} = 600\,......\left( {\text{i}} \right) \cr} $$
For path $$BC,$$
$$\eqalign{ & dQ = dU + dW \cr & \therefore 200 = {U_C} - {U_B} + {p_B}\left( {{V_C} - {V_B}} \right)\,\,\left( {{\text{as}}\,BC\,{\text{is}}\,{\text{isobaric}}\,{\text{process}}} \right) \cr & \therefore 200 = {U_C} - {U_B} + 8 \times {10^4}\left( {5 \times {{10}^{ - 3}} - 2 \times {{10}^{ - 3}}} \right) \cr & {\text{or}}\,\,200 = {U_C} - {U_B} + 240 \cr & \therefore {U_C} - {U_B} = - 40\;\,J\,......\left( {{\text{ii}}} \right) \cr} $$
For $$AC,$$ change in internal energy can be calculated by adding Eqs. (i) and (ii),
$$\eqalign{ & = {U_C} - {U_A} = {U_C} - {U_B} + {U_B} - {U_A} \cr & = - 40 + 600 = 560\,J \cr} $$
For path $$AB,$$ applying first law of thermodynamics
$$\eqalign{ & dQ = dU + dW\,\,{\text{or}}\,\,dQ = {U_B} - {U_A} + dW \cr & {\text{or}}\,\,600 = {U_B} - {U_A} + 0\,\,\left( {{\text{for}}\,{\text{isochoric}}\,{\text{process}}\,dV = 0\,\,{\text{so,}}\,\,dW = 0} \right) \cr & {U_B} - {U_A} = 600\,......\left( {\text{i}} \right) \cr} $$
For path $$BC,$$
$$\eqalign{ & dQ = dU + dW \cr & \therefore 200 = {U_C} - {U_B} + {p_B}\left( {{V_C} - {V_B}} \right)\,\,\left( {{\text{as}}\,BC\,{\text{is}}\,{\text{isobaric}}\,{\text{process}}} \right) \cr & \therefore 200 = {U_C} - {U_B} + 8 \times {10^4}\left( {5 \times {{10}^{ - 3}} - 2 \times {{10}^{ - 3}}} \right) \cr & {\text{or}}\,\,200 = {U_C} - {U_B} + 240 \cr & \therefore {U_C} - {U_B} = - 40\;\,J\,......\left( {{\text{ii}}} \right) \cr} $$
For $$AC,$$ change in internal energy can be calculated by adding Eqs. (i) and (ii),
$$\eqalign{ & = {U_C} - {U_A} = {U_C} - {U_B} + {U_B} - {U_A} \cr & = - 40 + 600 = 560\,J \cr} $$
