One mole of ice is converted into water at 273 K. The entropies of $\;{{\rm{H}}_2}{\rm{O}}$ (S) and $\;{{\rm{H}}_2}{\rm{O}}$ (l)are 38.20 and 60.01 J ${\rm{mo}}{{\rm{l}}^{ - 1}}{{\rm{K}}^{ - 1}}$ respectively. The enthalpy change for the conversion is:

$5954{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$59.54{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$595.4{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

$320.6{\rm{Jmo}}{{\rm{l}}^{ - 1}}$

The standard molar entropy of an ideal gas $\left(\gamma=\frac{4}{3}\right)$ is $2.5 \mathrm{~cal} / \mathrm{K}-\mathrm{mol}$ at $25^{\circ} \mathrm{C}$ and $1 \mathrm{~bar}.$ The standard molar entropy of the gas at $323^{\circ} \mathrm{C}$ and 1 bar is $(\ln 2=0.7)$

$5.6 \mathrm{~cal} / \mathrm{K}-\mathrm{mol}$

$\Delta_r H^{\circ}$ of $\mathrm{CS}_2$ is $117.36 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\Delta_{\mathrm{f}} \mathrm{H}^{\circ}$ of $\mathrm{C}(\mathrm{g})$ is $716.682 \mathrm{~kJ} \mathrm{~mol}^{-1}$ $\Delta_f H^{\circ}$ of $\mathrm{S}(g)$ is $278.805 \mathrm{~kJ} \mathrm{~mol}^{-1}$ Thus, $(\mathrm{C}=\mathrm{S})$ bond enthalpy in $\mathrm{CS}_2$ is

$1156.93 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The vapour pressure of benzene is $0.153 \times 10^5 \mathrm{Nm}^{-2}$ at $303 \mathrm{~K}$ and $0.520 \times 10^5 \mathrm{Nm}^{-2}$ at $333 \mathrm{~K}$. The mean latent heat of evaporation of benzene over this temperature range is

$48.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$31.8 \mathrm{~kJ} \mathrm{~mol}^{-1}$

$40.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$

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JEE • CHEMISTRY • THERMODYNAMICS

Thermodynamics practice · Hard

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Showing 1-30 of 92 questions

1 Hard

1 of 92

Analyzing +4 | -1

Consider following reactions
I:

\ \mathrm{C}(\mathrm{s})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}(\mathrm{g}), \Delta \mathrm{H}_1=\mathrm{a}

II:

\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})-\mathrm{CO}_2(\mathrm{~g}), \Delta \mathrm{H}_2=\mathrm{b}

III:

\mathrm{C}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})-2 \mathrm{CO}(\mathrm{g}) ; \Delta \mathrm{H}_3=\mathrm{c}

Select the correct statement

View answer and explanation

Correct option: D

All the above are correct statements

Option D is correct because all the given statements are correct based on Hess's Law and the definitions of heat of formation and combustion.

2 Hard

2 of 92

Application +4 | -1

When

0.18 \mathrm{~g}

of glucose was burnt in a bomb calorimeter, the temperature rose by

4 \mathrm{~K}.

The heat capacity of the calorimeter system is

\left(\Delta_{\mathrm{C}} \text{H}^{\circ}=-2.8\times 10^6 \mathrm{~J} / \mathrm{mol} \text{ glucose}\right)

View answer and explanation

Correct option: A

700 \mathrm{~J} / \mathrm{K}

Option A (

700 \mathrm{~J} / \mathrm{K}

) is correct because it is the heat capacity calculated using the heat released by the combustion of glucose and the temperature rise.

3 Hard

3 of 92

Application +4 | -1

A sample of oxygen gas expands its volume from

3 \mathrm{~L}

5 \mathrm{~L}

against a constant pressure of

3 \mathrm{~atm}.

If the work done during expansion is used to heat

10 \mathrm{~mol}

of water initially present at

290 \mathrm{~K},

its final temperature will be

(Specific heat capacity of water

=4.18 \mathrm{~JK}^{-1} \mathrm{~g}^{-1}

)

View answer and explanation

Correct option: B

290.8 \mathrm{~K}

Option B (

290.8 \mathrm{~K}

) is correct because it accurately calculates the final temperature of the water after accounting for the work done by the expanding oxygen gas.

4 Hard

4 of 92

Application +4 | -1

One mole of ice is converted into water at 273 K. The entropies of

\;{\mathrm{H}_2}\mathrm{O}

(S) and

\;{\mathrm{H}_2}\mathrm{O}

(l)are 38.20 and 60.01 J

\mathrm{mo}{\mathrm{l}^{ - 1}}{\mathrm{K}^{ - 1}}

respectively. The enthalpy change for the conversion is:

View answer and explanation

Correct option: B

5954\mathrm{Jmo}{\mathrm{l}^{ - 1}}

Option B (

5954\mathrm{Jmo}{\mathrm{l}^{ - 1}}

) is correct because it is the enthalpy change calculated using the formula ∆H=T∆S at equilibrium.

5 Hard

5 of 92

Application +4 | -1

In the reaction

\mathrm{CS}_2({l})+3 \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{CO}_2(\mathrm{g})+2 \mathrm{SO}_2(\mathrm{g}) ; \Delta H= -265 \text{ kcal}.

The enthalpies of formation of

\mathrm{CO}_2

and

\mathrm{SO}_2

are both negative and are in the ratio

4: 3.

The enthalpy of formation of

\mathrm{CS}_2

+26 \mathrm{~kcal} \mathrm{~mol}^{-1}.

Calculate the enthalpy of formation of

\mathrm{SO}_2.

View answer and explanation

Correct option: D

-71.7 \mathrm{~kcal} \mathrm{~mol}^{-1}

Option D (

-71.7 \mathrm{~kcal} \mathrm{~mol}^{-1}

) is correct because it is calculated using Hess's law and the given enthalpy of reaction and formation of other compounds.

6 Hard

6 of 92

Analyzing +4 | -1

Calculate the enthalpy of formation of

{I}_2 {O}_5(\mathrm{s})

from the following data.

(i)

{I}_2 {O}_5(\mathrm{s})+{H}_2{O}({l}) \rightarrow 2 {HIO}_3\text{(aq)}; \Delta H=+4.0 \mathrm{~kJ}

(ii)

{KI} \text{(aq)} +3 {HClO}\text{(aq)} \rightarrow {HIO}_3 \text{(aq)} +2 {HCl}\text{(aq)}+{KCl}(\mathrm{aq}); \Delta H=-322.0 \mathrm{~kJ}

(iii)

{NaOH}(\mathrm{aq})+{HClO}(\mathrm{aq}) \rightarrow {NaOCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-44.0 \mathrm{~kJ}

(iv)

{NaOH}(\mathrm{aq})+{HCl}(\mathrm{aq}) \rightarrow {NaCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-57.0 \mathrm{~kJ}

(v)

2 {NaOH}(\mathrm{aq})+{Cl}_2(\mathrm{g}) \rightarrow{NaOCl}(\mathrm{aq})+ {NaCl}(\mathrm{aq})+{H}_2 {O}({l}); \Delta H=-100.0 \mathrm{~kJ}

(vi)

2 {KI}(\mathrm{aq})+{Cl}_2(\mathrm{g}) \rightarrow 2 {KCl}(\mathrm{aq})+{I}_2(\mathrm{s}); \Delta H=-224.0 \mathrm{~kJ}

(vii)

{H}_2(\mathrm{g})+1 / 2 \, {O}_2(\mathrm{g}) \rightarrow {H}_2 {O}({l}); \Delta H=-285.0 \mathrm{~kJ}

(viii)

1 / 2 \,{H}_2(\mathrm{g})+1 / 2 \, {Cl}_2(\mathrm{g}) \rightarrow {HCl}(\mathrm{g}); \Delta H= -92.0 \mathrm{~kJ}

(ix)

{HCl}(\mathrm{g})+\mathrm{aq} \rightarrow {HCl}(\mathrm{aq}); \Delta H=-75.0 \mathrm{~kJ}

View answer and explanation

Correct option: A

-169.0 \mathrm{~kJ}

Option A (-169.0 kJ) is correct because the enthalpy of formation of I2O5(s) is calculated by combining the given reactions using Hess's Law.

7 Hard

7 of 92

Analyzing +4 | -1

A solution of

500 \mathrm{~ml}

2 \mathrm{~M}-{KOH}

is added to

500 \mathrm{~ml}

2 \mathrm{~M}-{HCl}

and the mixture is well shaken. The rise in temperature

T_1

is noted. The experiment is again performed using

250 \mathrm{~ml}

{KOH}

solution and

500 \mathrm{~ml}

{HCl}

solution and rise in temperature

T_2

is again noted. Which of the following is correct?

View answer and explanation

Correct option: D

T_1

is 1.5 times as larger as

T_2

Option D (T1 is 1.5 times as larger as T2) is correct because the heat released in the first experiment is 1.5 times that of the second experiment.

8 Hard

8 of 92

Understanding +4 | -1

During complete combustion of one mole of butane,

2658 \mathrm{~kJ}

of heat is released. The thermochemical reaction for above change is

View answer and explanation

Correct option: C

\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

The correct answer is C (

\mathrm{C}_4 \mathrm{H_{10}(g})+\frac{13}{2} \mathrm{O_2(g}) \rightarrow 4 \mathrm{CO_2(g})+ 5 \mathrm{H}_2 \mathrm{O}(l) ; \Delta_C H =-2658.0 \mathrm{~kJ} \mathrm{~mol}^{-1}

) because it represents the complete combustion of one mole of butane with the correct stoichiometric coefficients and enthalpy change.

9 Hard

9 of 92

Application +4 | -1

If values of

\Delta H^{\circ} \,\&\, \Delta S^{\circ}

for a process/reaction are

77.2 \mathrm{~kJ} \,\&\, 48 \mathrm{~J} / \mathrm{k}

respectively. Then find value of

\log \frac{1}{k}

. Given: Temp is

300 \mathrm{~K}

View answer and explanation

Correct option: A

Option A (11) is correct because it involves calculating the Gibbs free energy change and then using it to find the equilibrium constant, finally calculating the log of the inverse of the equilibrium constant.

10 Hard

10 of 92

Application +4 | -1

One mole of a liquid (

1 \mathrm{~bar}, 100 \mathrm{~ml}

) is taken in an adiabatic container and the pressure increases steeply to

100 \mathrm{~bar}.

Then at a constant pressure of

100 \mathrm{~bar},

volume decrease by

1 \mathrm{~ml}

. Find

\Delta \mathrm{H}

View answer and explanation

Correct option: A

990 J

Option A (990 J) is correct because it is the enthalpy change calculated from the given pressure and volume changes in the adiabatic container.

11 Hard

11 of 92

Application +4 | -1

The bond energies of

\mathrm{C} \equiv \mathrm{C}, \mathrm{C}-\mathrm{H}, \mathrm{H}-\mathrm{H}

and

\mathrm{C}=\mathrm{C}

are

198,98,103,145 \mathrm{~kcal}

respectively. The enthalpy change of the reaction

\mathrm{CH} \equiv \mathrm{CH}+\mathrm{H}_2 \rightarrow \mathrm{CH}_2=\mathrm{CH}_2

View answer and explanation

Correct option: D

-40 \mathrm{~kcal}

Option D (-40 kcal) is correct because the enthalpy change is calculated by subtracting the bond energies of products from the bond energies of reactants.

12 Hard

12 of 92

Analyzing +4 | -1

Statement-1: Work done (W) by the system during isobaric reversible and irreversible processes from same initial state to same final state is equal.
Statement-2: As the pressure is constant so

\mathrm{W}=-\mathrm{P}(\Delta \mathrm{V})

for both the processes.

View answer and explanation

Correct option: A

Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1

The correct answer is A because work done in an isobaric process is given by

W = -P\Delta V

, and since the initial and final states are the same,

\Delta V

is the same for both reversible and irreversible processes.

13 Hard

13 of 92

Application +4 | -1

The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0℃ is:

View answer and explanation

Correct option: A

5.260 cal/(mol K)

Option A (5.260 cal/(mol K)) is correct because it is calculated by dividing the enthalpy of fusion by the temperature in Kelvin.

14 Hard

14 of 92

Application +4 | -1

S + \frac{3}{2}{O_2} \to S{O_3} + 2x\,\,kcal.....\left( i \right)\,\,\,\,S{O_2} + \frac{1}{2}{O_2} \to S{O_3} + y\,\,kcal.....\left( {ii} \right)

Find out the Enthalpy of formation of

S{O_2}

View answer and explanation

Correct option: A

y – 2x

The correct answer is A (y – 2x) because it is derived by manipulating the given equations to find the enthalpy of formation of SO2.

15 Hard

15 of 92

Application +4 | -1

For silver,

{C_p}(J{K^{ - 1}}\,mo{l^{ - 1}})\, = \,23\, + \,0.01\,T

. If the temperature

(T)

of 3 moles of silver is raised from 300 K to 1000 K at 1 atm, pressure, the value of

\Delta H

will be close to

View answer and explanation

Correct option: A

62 kJ

Option A (62 kJ) is correct because it involves using Kirchoff's relation to calculate the change in enthalpy by integrating the heat capacity equation over the given temperature range.

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16 Hard

16 of 92

Application +4 | -1

The dissolution of

{CaCl}_2 \cdot 6 {H}_2 {O}

in a large volume of water is endothermic to the extent of

3.5 \mathrm{~kcal/mol}.

For the reaction,

{CaCl}_2(\mathrm{s})+6 {H}_2 {O}({l}) \rightarrow{CaCl}_2 \cdot 6 {H}_2 {O}(\mathrm{s}) ; \Delta H

-23.2 \mathrm{~kcal}.

The heat of solution of anhydrous

{CaCl}_2

in large quantity of water will be

View answer and explanation

Correct option: B

-19.7 \mathrm{~kcal} \mathrm{~mol}^{-1}

The correct answer is B (-19.7 kcal mol⁻¹) because it is calculated by using Hess's law to relate the enthalpy changes of the given reactions.

17 Hard

17 of 92

Application +4 | -1

56 \mathrm{~g}

of iron reacts with

\mathrm{HCl}

in a closed vessel of fixed volume. Again, same amount of

\mathrm{Fe}

is made to react in an open beaker at

27^{\circ} \mathrm{C}

. The work done (in cal) in both the cases will be respectively

View answer and explanation

Correct option: D

0,-600

Option D (0, -600) is correct because in a closed vessel with fixed volume, no work is done, while in an open beaker, work is done due to the change in the number of moles of gas.

18 Hard

18 of 92

Application +4 | -1

A diatomic gas at pressure

\mathrm{P}

, compressed adiabatically to half of its volume, what is the final pressure?

View answer and explanation

Correct option: A

(2)^{1.4} \mathrm{P}

Option A ((2)¹.4 P) is correct because it is the final pressure calculated using the adiabatic compression formula.

19 Hard

19 of 92

Application +4 | -1

For the reaction,

\mathrm{CaO}(s)+\mathrm{SO}_3(g) \longrightarrow \mathrm{CaSO}_4(s) ; \Delta G^{\circ}=-345.0 \mathrm{~kJ}

\Delta G

for the reaction at

298 \mathrm{~K}

when the pressure of

\mathrm{SO}_3

0.5 \mathrm{~atm}

View answer and explanation

Correct option: D

-343.3 \mathrm{~kJ}

Option D (-343.3 kJ) is correct because it is calculated using the Gibbs free energy equation considering the non-standard conditions with the given pressure of SO3.

20 Hard

20 of 92

Application +4 | -1

The standard molar entropy of an ideal gas

\left(\gamma=\frac{4}{3}\right)

2.5 \mathrm{~cal} / \mathrm{K}-\mathrm{mol}

25^{\circ} \mathrm{C}

and

1 \mathrm{~bar}.

The standard molar entropy of the gas at

323^{\circ} \mathrm{C}

and 1 bar is

(\ln 2=0.7)

View answer and explanation

Correct option: D

8.1 \mathrm{~cal} / \mathrm{K}-\mathrm{mol}

Option D (

8.1 \mathrm{~cal} / \mathrm{K}-\mathrm{mol}

) is correct because it is calculated using the formula for entropy change with temperature at constant pressure.

21 Hard

21 of 92

Application +4 | -1

Calculate the lattice energy forthe reaction,

\mathrm{Li}^{+}(\mathrm{g})+\mathrm{Cl}^{-}(\mathrm{g}) \longrightarrow \mathrm{LiCl}(\mathrm{s})

From the following data:

\Delta_{\mathrm{s}} \mathrm{H}^{\circ}(\mathrm{Li})=160.67 \mathrm{~kJ} \mathrm{~mol}^{-1} ; \quad \frac{1}{2} \Delta_{\mathrm{a}} \mathrm{H}^{\circ}\left(\mathrm{Cl}_2\right)=122.17 \mathrm{~kJ} \mathrm{~mol}^{-1}

\Delta_{\mathrm{i}} \mathrm{H}^{\circ}(\mathrm{Li})=520.07 \mathrm{~kJ} \mathrm{~mol}^{-1}

;

\quad

\Delta_{\mathrm{eg}} \mathrm{H}^{\circ}(\mathrm{Cl})=-365.26 \mathrm{~kJ} \mathrm{~mol}^{-1}

\text { and } \Delta_{\mathrm{f}} \mathrm{H}^{\circ}(\mathrm{LiCl})=-401.66 \mathrm{~kJ} \mathrm{~mol}^{-1}

View answer and explanation

Correct option: A

-839.31 \mathrm{~kJ~mol}^{-1}

Option A (

-839.31 \mathrm{~kJ~mol}^{-1}

) is correct because it is the calculated lattice energy using the given enthalpy values and applying Hess's law.

22 Hard

22 of 92

Analyzing +4 | -1

The data below refers to gas phase reaction at constant pressure at

25^{\circ} \mathrm{C}.

\begin{aligned} & \text{(a) } \mathrm{CH}_3-\mathrm{CH}_3 \rightarrow \mathrm{CH}_3-\mathrm{CH}_2+\mathrm{H}; \Delta H_1=+420 \mathrm{~kJ} \mathrm{~mol}^{-1} \\ & \text{(b) } \mathrm{CH}_3-\mathrm{CH}_2 \rightarrow \mathrm{CH}_2=\mathrm{CH}_2+\mathrm{H}; \Delta H_2=+168 \mathrm{~kJ} \mathrm{~mol}^{-1} \end{aligned}

From these data, the enthalpy change

\Delta H

for the reaction

2 \mathrm{CH}_3-\mathrm{CH}_2 \rightarrow \mathrm{CH}_3-\mathrm{CH}_3+\mathrm{CH}_2=\mathrm{CH}_2

View answer and explanation

Correct option: C

-252 \mathrm{~kJ}

Option C (

-252 \mathrm{~kJ}

) is correct because the enthalpy change is calculated by subtracting the enthalpy of reaction (a) from reaction (b).

23 Hard

23 of 92

Application +4 | -1

1 mole of an ideal gas undergoes reversible isothermal expansion from an initial volume

V_1

to a final volume

10 V_1

and does

10 {~kJ}

of work. The initial pressure was

1 \times 10^7 {~Pa}

. Calculate

V_1

View answer and explanation

Correct option: A

4.34\times10^{-4}\,m^{3}

Option A (

4.34\times10^{-4}\,m^{3}

) is correct because it is the initial volume calculated using the isothermal reversible work formula and the ideal gas law.

24 Hard

24 of 92

Application +4 | -1

The enthalpy of atomization of graphite is

698.6 \, \mathrm{kJ} / \mathrm{mol}

and the mean bond enthalpy of

\mathrm{C}-\mathrm{C}

bond in diamond is

348.4 \mathrm{~kJ} / \mathrm{mol}.

The enthalpy of conversion of graphite into diamond is

View answer and explanation

Correct option: B

+1.8 \mathrm{~kJ} / \mathrm{mol}

Option B (+1.8 kJ/mol) is correct because the enthalpy of conversion is calculated by subtracting twice the C-C bond enthalpy in diamond from the enthalpy of atomization of graphite.

25 Hard

25 of 92

Application +4 | -1

1 mole of gas occupying 3 L volume is expanded against a constant external pressure of 1 atm to a volume of 15 litre. The work done by the system is:

View answer and explanation

Correct option: A

1.215 \times {10^3}\mathrm{J}

Option A (

1.215 \times {10^3}\mathrm{J}

) is correct because the work done is calculated using the formula

W = -P_{ext} \Delta V

and converting litre atm to Joules.

26 Hard

26 of 92

Application +4 | -1

\Delta_r H^{\circ}

\mathrm{CS}_2

117.36 \mathrm{~kJ} \mathrm{~mol}^{-1}

\Delta_{\mathrm{f}} \mathrm{H}^{\circ}

\mathrm{C}(\mathrm{g})

716.682 \mathrm{~kJ} \mathrm{~mol}^{-1}

\Delta_f H^{\circ}

\mathrm{S}(g)

278.805 \mathrm{~kJ} \mathrm{~mol}^{-1}

Thus,

(\mathrm{C}=\mathrm{S})

bond enthalpy in

\mathrm{CS}_2

View answer and explanation

Correct option: B

578.47 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option B (

578.47 \mathrm{~kJ} \mathrm{~mol}^{-1}

) is correct because it involves calculating the bond enthalpy of C=S in CS2 using enthalpy of formation and atomization energies.

27 Hard

27 of 92

Application +4 | -1

The vapour pressure of benzene is

0.153 \times 10^5 \mathrm{Nm}^{-2}

303 \mathrm{~K}

and

0.520 \times 10^5 \mathrm{Nm}^{-2}

333 \mathrm{~K}

. The mean latent heat of evaporation of benzene over this temperature range is

View answer and explanation

Correct option: D

34.2 \mathrm{~kJ} \mathrm{~mol}^{-1}

Option D (

34.2 \mathrm{~kJ} \mathrm{~mol}^{-1}

) is correct because it's the result of applying the Clausius-Clapeyron equation using the given vapor pressures and temperatures.

28 Hard

28 of 92

Application +4 | -1

\mathrm{H}_2(\mathrm{g})+\frac{1}{2} \mathrm{O}_2(\mathrm{g}) \rightarrow \mathrm{H}_2 \mathrm{O}({l})

B.E.

(\mathrm{H}-\mathrm{H})=x_1;

B.E.

(\mathrm{O}=\mathrm{O})=x_2;

B.E.

(\mathrm{O}-\mathrm{H})=x_3

Latent heat of vaporization of water liquid into water vapor

=x_4,

then

\Delta H_{\mathrm{f}}

(heat of formation of liquid water) is

View answer and explanation

Correct option: C

x_1+\frac{x_2}{2}-2 x_3-x_4

Option C (

x_1+\frac{x_2}{2}-2 x_3-x_4

) is correct because it represents the heat of formation of liquid water considering bond energies and the latent heat of vaporization.

29 Hard

29 of 92

Application +4 | -1

One mole of an ideal monoatomic gas is heated in a process

P V^{5 / 2}=

constant. By what amount heat is absorbed in the process in

26^{\circ} \mathrm{C}

rise in temperature?

View answer and explanation

Correct option: B

180 \mathrm{~J}

Option B (180 J) is correct because the heat absorbed is calculated using the molar heat capacity for a polytropic process and the given temperature change.

30 Hard

30 of 92

Application +4 | -1

With what minimum pressure (in

\mathrm{kPa}

), a given volume of an ideal gas

\left(C_{p, m}=7 / 2 R\right)

, originally at

400 \mathrm{~K}

and

100 \mathrm{kPa}

pressure can be compressed irreversibly adiabatically in order to raise its temperature to

600 \mathrm{~K}

View answer and explanation

Correct option: B

275 \mathrm{~kPa}

Option B (

275 \mathrm{~kPa}

) is correct because it's derived from the adiabatic irreversible compression formula considering the given conditions and

C_{p,m}

value.

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