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Work Done in Isothermal Reversible Process

The standard ...

Question

The standard state Gibbs free energies of formation of $C (g r a p h i t e)$ and $C (d i a m o n d)$ at T = 298 K are

$Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}$
$Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 k J m o l^{- 1}$

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by $2 \times 10^{- 6} m^{3} m o l^{- 1}$ . If $C (g r a p h i t e)$ is converted to $C (d i a m o n d)$ isothermally at T = 298 K, the pressure at which $C (g r a p h i t e)$ is in equilibrium with $C (d i a m o n d)$ is:
[Useful information: $1 J = 1 k g m^{2} s^{- 2}$ ; $1 P a = 1 k g m^{- 1} s^{- 2}$ ; $1 b a r = 10^{5} P a$ ]

14501 bar

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58001 bar

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1450 bar

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29001 bar

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Solution

The correct option is A 14501 bar
$d G = V d P - S d T$
at constant temperature (298K),
$S d T = 0$
$d G = V d P$
$\int_{1}^{P} d G$ = $\int_{1}^{P} V d P$
So,
$G - G^{0}$ = V (P-1) [Since solids are involved therefore V is almost constant]
So, we have
$Δ_{r} G = [G_{d i a m o n d}^{0} + V_{d} (P - 1)] - [G_{g r a p h i t e}^{0} + V_{g} (P - 1)]$
0= $2.9 \times 10^{3} + (P - 1) 10^{5} (- 2 \times) 10^{- 6}$
$P_{i}$ = 1 bar
$P = 14501 b a r$

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Similar questions

Q. The standard state Gibbs free energies of formation of C(graphite) and C(diamond) at T = 298K are

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 K J m o l^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

. If C(graphite) is converted to C(diamond) isothermally at T = 298K, the pressure at which C(graphite) is in equilibrium with C(diamond), is
[Useful information:

1 J = 1 k g m^{2} s^{- 2}

1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a

Q. The standard state Gibbs free energies of formation of

C_{(g r a p h i t e)}

and

C_{(d i a m o n d)}

T = 298 K

are

Δ_{f} G^{\circ} C_{(g r a p h i t e)} = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} C_{(d i a m o n d)} = 2.9 k J m o l^{- 1}

The standard state means that the pressure should be

1 b a r

, and substance should be pure at a given temperature. The conversion of graphite

C_{(g r a p h i t e)}

to diamond

C_{(d i a m o n d)}

reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

. If

C_{(g r a p h i t e)}

is converted to

C_{(d i a m o n d)}

isothermally at

T = 298 K

, the pressure at which

C_{(g r a p h i t e)}

is in equilibrium with

C_{(d i a m o n d)}

, is?

[Useful information:

1 J = 1 k g m^{2} s^{- 2}; 1 P a = 1 k g m^{- 1} s^{- 2}; 1 b a r = 10^{5} P a]

Q. The standard state Gibbs free energies of formation of

C (g r a p h i t e)

and

C (d i a m o n d)

at T = 298 K are

Δ_{f} G^{\circ} [C (g r a p h i t e)] = 0 k J m o l^{- 1}

Δ_{f} G^{\circ} [C (d i a m o n d)] = 2.9 k J m o l^{- 1}

The standard state means that the pressure should be 1 bar, and substance should be pure at a given temperature. The conversion of graphite [C(graphite)] to diamond [C(diamond)] reduces its volume by

2 \times 10^{- 6} m^{3} m o l^{- 1}

. If

C (g r a p h i t e)

is converted to

C (d i a m o n d)

isothermally at T = 298 K, the pressure at which

C (g r a p h i t e)

is in equilibrium with

C (d i a m o n d)

is:
[Useful information:

1 J = 1 k g m^{2} s^{- 2}

;

1 P a = 1 k g m^{- 1} s^{- 2}

;

1 b a r = 10^{5} P a

]

Q. The enthalpy of transition of C(diamond) to C(graphite) is (Given that the enthalpies of combustion of C(diamond) and C(graphite) are – 393.5 and – 395.4 kJ

m o l^{- 1}

respectively)

Q. The enthalpy of transition of C(diamond) to C(graphite) is?
Given that the enthalpies of combustion of C(diamond) and C(graphite) are – 393.5 and – 395.4 kJ

m o l^{- 1}

respectively)

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