For isothermal expanison in case of an ideal gas:-
For ideal gas isothermal expansion
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For isothermal expanison in case of an ideal gas:-
For ideal gas isothermal expansion
At 25C, for the process H2O(l)H2O(g);
is 8.6 kJ. The vapour pressure of water at this temperature is nearly:-
At equilibrium = -RTlnKeq
Keq = /P and P= 1 bar
What will be the standard enthalpy of reaction for the following reaction using the listed enthalpies of reaction:-
3Co(s) + 2O2(g)Co3O4(s); =?
2Co(s) + O2(g) 2CoO(s) = -475.8 KJ
6CoO(s) + O2(g) 2Co3O4(s) = -355.0 KJ
= -713.7 - 177.5 = -891.2 KJ
For given following equations and values determine the enthalpy of reaction at 298 K for the reaction
C2H4(g) + 6F2(g) 2CF4(g) + 4HF(g)
H2(g) + F2(g) 2HF(g) = -537 kJ
C(s) + 2F2(g) CF4(g) =-680 kJ
2C(s) + 2H2(g) C2H4(g) = 52 kJ
If (C2H4) and (C2H6) are x1 and x2 kcal/mol then heat of hydrogenation of C2H4 is :-
3. Target reaction C2H4 + H2 C2H6
Given reaction
(i) 2C + 2H2 C2H4 =x1
(ii) 2C + 3H2C2H6 = x2
(ii)-(i) = x2 - x1
For oxidation of iron, 4Fe(s) + 3O2(g) 2Fe2O3(s), =-1648x103 J mol-1 entropy change is:
-549.4 J k-1mol-1 at 298 K:-
The reaction is
18 g of ice is converted into water at 0C and 1 atm. The entropy of H2O(s) and H2O(l) are 38.2 and 60 J K-1 mol-1 respectively. the enthalpy for this conversion will be
= 60-38.2=21.8 JK-1 mol-1
... =21.8 x 273 = 5951.4 J/mol
For a given reaction, H =35.5 kJmol-1 and S = 83.6JK-1 mol-1. The reaction is spontaneous at: (Assume that H and S do not vary with temperature)
(b) According to Gibbs-Helmholtz equation, Gibbs energy (G) = H - TS
where, H = Enthalpy change
S = Entropy change
T = Temperature
For a reaction to be spontaneous G <0.
Gibbs -Helmholtz equation becomes.
G = H -TS<0
or, H < TS
T > H / S
=35.5 KJmol-1/ 83.6 JK-1mol-1
=35.5 x 1000J mol-1 / 83.6JK-1mol-1
=425 K
T>425K
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy U of the gas in joules will be
(c) Key concept According to first law of thermodynamics,
U = q + w
where, U =internal energy
q = heat absorbed or evolved, w = work done.
Also, work done against constant external pressure (irreversible process).
W = -Pext V.
Work done in irreversible process,
w = -Pext V = - pext (V2 - V1)
= -2.5 atm (4.5 L - 2.5 L)
= - 5 L atm = - 5 x 101.3 J
= - 505 J
Since, the system is well insulated, q =0
U = w = - 505 J
Hence, change in internal energy, U of the gas is - 505 J.
For a sample of perfect gas when its pressure is changed isothermally from pi to pf, the entropy change is given by
(b) Entropy change is given as,
S = nCpln(Tf/Ti) + nRln(pi/pf) ....(i)
For isothermal process, Ti = Tf
... nCpln(Tf/Ti) =0 [ln1 = 0]
From Eq (i) S = nRln(pi/pf)
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