For 1 mole of an ideal gas,
PV=RT
P=4.1 atm
V=2lt.
PV=RT
T=\(\frac{PV}{R}=\frac{4.1atm\times2lit\times1mole}{0.082 lit atm K^{-1}mol^{-1}}\)
=100 K
\(\Delta S=\frac{q}{T}\)
\(\Delta S=\frac{3710J}{100K}=37.1 JK^{-1}\)
ΔS of expansion = 37.1 JK^-1.

\(CO+\frac { 1 }{ 2 } { O }_{ 2 }\longrightarrow { CO }_{ 2 }\)
\({ \triangle }G_{ (reaction) }^{ 0 }={ \sum { G } }_{ f(products) }^{ 0 }-{ \sum { G } }_{ f(reactants) }^{ 0 }\)
\({ \triangle G }_{ (reaction) }^{ 0 }=\left[ { G }_{ { CO }_{ 2 } }^{ 0 } \right] -\left[ { G }_{ CO }^{ 0 }+\frac { 1 }{ 2 } { G }_{ { O }_{ 2 } }^{ 0 } \right] \)
\({ \triangle G }_{ (reaction) }^{ 0 }\) =-394.4+[137.2+0]
\({ \triangle G }_{ (reaction) }^{ 0 }\) =-257.2 kJ mol^-1
\({ \triangle G }_{ (reaction) }^{ 0 }\) of a reaction at a given temperature is negative hence the reaction is spontaneous.

ΔH_t = 2090 Jmol^-1
T_t = 13+273 = 286K
ΔS_t = \(\frac { { \triangle H }_{ t } }{ { T }_{ t } } \)
ΔS_t = \(\frac { 2090 }{ 286 } \)
ΔS_t = 7.307 JK^-1mol^-1​​​​​​​

C+O₂ ➝ CO₂     \(\Delta { S }_{ f }^{ o }\)=?
\(\Delta { S }_{ f }^{ o }\), CO₂ = \({ \Sigma S }_{ Compound }^{ o }-{ \Sigma S }_{ elements }^{ o }\)
=218.8(8.74+205.60)
\(\Delta { S }_{ f }^{ o }\), CO₂ =4.46 Jk^-1

From the definition of enthalpy
H = U+PV ..... (1)
for 1 mole of an ideal gas
PV = nRT .....(2)
By substituting (2) in (1)
H = U + nRT ..... (3)
Differentiating the above equation with respect to T,
\(\frac { \partial H }{ \partial T } =\frac { \partial U }{ \partial T } +nR\frac { \partial T }{ \partial T } \)
C_p= C_v+nR(1)
C_p-C_v=nR \(\left[ \because \left( \frac { \partial H }{ \partial T } \right) _{ p }={ C }_{ p }\ and\ \left( \frac { \partial U }{ \partial T } \right) _{ v }={ C }_{ v } \right] \)
At constant pressure processes, a system has to do work against the surroundings. Hence, the system would require more heat to effect a given temperature rise than at constant volume, so C_p is always greater than C_v

The standard enthalpy of a reaction is the enthalpy change for a reaction when all the participating substances are present in their standard states. Standard conditions are denoted by adding the superscript 0 to the symbol (ΔH⁰).

ΔU=q-w
= 100 - 50 = 50J
Since work is done by the system, it is -ve.

No transfer of heat or work is observed in an isolated system.
∴ΔU=q+w
ΔU=0+0=0.

Zeroth law of thermodynamics states that 'If two systems at different temperatures are separately in thermal equilibrium with a third one, then they tend to be in thermal equilibrium with themselves'.

ΔG⁰ = - 2.303 RT log k_c
= - 2.303 x 8.314 x 298 log (1.8 x 10^-7)
= - 38484 J mol^-1
= - 38.484 kJ mol-1

Δ H⁰= ΔU⁰ + RT(Δn)
= -10.5 + 8.314 x 10^-3 x 298 x (-1)
= -12.978 kJ
We know, ΔG⁰= Δ H⁰ - TΔS⁰ 
= -12.978 - 298 (- 44.1 x 10^-3)
= 0.164 kJ
Hence, the reaction is non spontaneous.

Work done in free expansion or work done against zero pressure is also zero.
w_exp= -P_ext ( \(\Delta\)U)
0 = w_exp=-P_ext (V₂-V₁)
0 = 0(10-2)
=0
Hence heat change, and work done is zero. 

\(\Delta S=\frac { { q }_{ rev } }{ T } =\frac { \Delta { H }_{ rev } }{ T } \)
Where \(\Delta\)H_rev is the enthalpy change at Temperature (T)