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A sample of argon gas at 1 atm pressure and $ 27 ^\circ C $ expands reversibly and adiabatically from $ 1.25 dm^3 to 2.50 dm^ 3 $ . The enthalpy change in this process will be……….$ [Cv.m. for argon is 12.48 jK^{–1} mol^{–1} ]$.
Find $ OG ^\circ $ and $ OH ^\circ $ for that the reaction $ CO(g) + O_2 (g) \rightarrow CO_2 (g) $ at 300 K respectively are, when the standard entropy change is $ – 0.094 kJ mol^{–1} K^{–1} $ . The standard Gibbs free energies of formation for $ CO_2 and CO are – 394.4 and – 137.2 kJ mol^{–1} $ , respectively.
To solve for $\Delta G^\circ$ and $\Delta H^\circ$, we use the Gibbs free energy change formula: $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$. Given: $T = 300\,K$, $\Delta S^\circ = -0.094\,kJ\,mol^{-1}\,K^{-1}$, $ ext{standard Gibbs free energies of formation: } \Delta G^\circ_{CO_2} = -394.4\,kJ\,mol^{-1}$ and $\Delta G^\circ_{CO} = -137.2\,kJ\,mol^{-1}$.\n\nFirst, calculate $\Delta G^\circ$ for the reaction: $\Delta G^\circ = (-394.4\,kJ\,mol^{-1}) - (-137.2\,kJ\,mol^{-1}) = -257.2\,kJ\,mol^{-1}$.\n\nNow, use the Gibbs free energy formula to find $\Delta H^\circ$: \n$-257.2\,kJ\,mol^{-1} = \Delta H^\circ - (300\,K)(-0.094\,kJ\,mol^{-1}\,K^{-1})$.\n\nSolving for $\Delta H^\circ$: \n$\Delta H^\circ = -257.2\,kJ\,mol^{-1} + 28.2\,kJ\,mol^{-1} = -285.4\,kJ\,mol^{-1}$.\n\nThus, the correct values are $\Delta G^\circ = -257.2\,kJ\,mol^{-1}$ and $\Delta H^\circ = -285.4\,kJ\,mol^{-1}$. The correct option is $o4$.
$ OH = 30 kJ mol^{–1} , OS = 75 J / k / mol $ . Find boiling temperature at 1 atm.
To find the boiling temperature at 1 atm, use the Gibbs free energy equation at equilibrium: $\Delta G = \Delta H - T\Delta S = 0$. \n\nGiven: $\Delta H = 30\,kJ\,mol^{-1}$ and $\Delta S = 75\,J\,mol^{-1}\,K^{-1} = 0.075\,kJ\,mol^{-1}\,K^{-1}$.\n\nSet $\Delta G$ to zero and solve for $T$: \n$0 = 30\,kJ\,mol^{-1} - T(0.075\,kJ\,mol^{-1}\,K^{-1})$.\n\n$T = \frac{30\,kJ\,mol^{-1}}{0.075\,kJ\,mol^{-1}\,K^{-1}} = 400\,K$.\n\nThus, the boiling temperature at 1 atm is 400 K. The correct option is $o1$.
Spontaneous adsorption of a gas on a solid surface is exothermic process because
Spontaneous adsorption of a gas on a solid surface is generally an exothermic process. During adsorption, gas molecules adhere to the surface, resulting in a decrease in disorder (entropy) of the gas molecules because they are more confined. This leads to a decrease in the entropy of the system.\n\nTherefore, the correct option is $o3$, stating that entropy decreases during spontaneous adsorption.
The ratio of P to V at any instant is constant and is equal to 1, for a monoatomic ideal gas under going a process. What is the molar heat capacity of the gas
From first law of Thermodynamics, $ \triangle E = q + w \Rightarrow nC_vdT = nCdT – PdV...... (1) $
Now according to process, P = V and according to ideal gas equation, PV = nRT We have, V2 = nRT $ On differentiating, 2VdV = nRdT and PdV = VdV = { nRdT \over 2 } $
So from first equation we have, $ nC_vdT = nCdT – { nRdT \over 2 } $ So, $ C_v = C – { R \over 2 } $ $ Hence C = { 4R \over 2 } $
The entropy values (in J K–1 mol–1) of H2 (g) = 130.6 Cl2(g) = 223 and HCl(g) = 186.7 at 298 K and 1 atmpressure are given. Then entropy change for the reaction.
A mixture of 2 mole of CO(g) and one mole of O2 in a closed vessel, is ignited to convert the carbon monoxide to carbon dioxide. If $ \triangle H and \triangle U$ are enthalpy and internal energy change. Then
For the reaction of one mole zinc dust with one sulphuric acid in a bomb calorimeter, $ \triangle U $ and w correspond to :
In a bomb calorimeter, the volume is constant, so no work is done (w = 0). For the reaction of zinc with sulfuric acid: $$ ext{Zn(s) + H}_2 ext{SO}_4(aq) ightarrow ext{ZnSO}_4(aq) + ext{H}_2(g) $$, the internal energy change $$ riangle U $$ is negative because the reaction releases energy (exothermic reaction). Hence, the correct option is: $$ riangle U < 0, w = 0 $$.
If the enthalpies of formation of $Al_2O_3 and Cr_2O_3$ are – 1596 kJ and – 1134 kJ respectively, then the value of OH for the reaction ; $ 2Al + Cr_2O_3 \rightarrow 2Cr + Al_2O_3$ is :
To find the change in enthalpy for the reaction, we use the enthalpies of formation of the reactants and products. The reaction given is: $$2Al + Cr_2O_3 ightarrow 2Cr + Al_2O_3$$ The enthalpy change of the reaction ($ riangle H$) can be calculated using the formula: $$ riangle H = ext{Sum of enthalpies of formation of products} - ext{Sum of enthalpies of formation of reactants}$$ For the given reaction: Products: $Al_2O_3$ with enthalpy of formation = -1596 kJ Reactants: $Cr_2O_3$ with enthalpy of formation = -1134 kJ So, $$ riangle H = [-1596] - [-1134] = -1596 + 1134 = -462 ext{ kJ}$$ Therefore, the correct answer is -462 kJ.
The internal energy change when a system goes from state A to B is 40 kJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy
The internal energy change ($ riangle U$) for a system depends only on the initial and final states and is independent of the path taken. Since the system returns to its initial state (A), the net change in internal energy is zero. This is a fundamental concept in thermodynamics: $$ riangle U_{ ext{net}} = riangle U_{ ext{A to B}} + riangle U_{ ext{B to A}} = 40 ext{ kJ} + (-40 ext{ kJ}) = 0$$ Hence, the net change in internal energy is zero.
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