Thermodynamics

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

A cylinder of fixed capacity of 44.8 litres contains helium gas at standard gas at standard temperature and pressure. The amount of heat needed to raise the temperature of gas in the cylinder by 20.0°C will be:

(Given gas constant R = 8.3 JK^-¹ -mol^-¹)

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

44.4 litres of He $\frac{44.8}{22.4} m o l e s$

$η = 2$ moles of He

Q = $η C_{v} Δ T$ (for fixed capacity (v = constant)

&nb

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

A gas has n degrees of freedom. The ratio of specific heat of gas at constant volume to the specific heat of gas at constant pressure will be:

2 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$C v = \frac{n \times R}{2}$

$C p = C v + R = \frac{(n + 2) R}{2}$

$\Rightarrow \frac{C v}{C P} = \frac{n}{(n + 2)}$

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8} .$ The value of x is

[R = universal gas constant]

5 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

$Δ Q = Δ U + Δ W$

$\Rightarrow Q = Δ U + \frac{Q}{5} \Rightarrow Δ U = \frac{4 Q}{5} = n C_{v} Δ T \Rightarrow \frac{4 Q}{5} = \frac{5 R}{2} Δ T \Rightarrow Δ T = \frac{8 Q}{25 R}$

$Q = n c Δ T = 1 \times C \times \frac{80}{25 R} \Rightarrow C = \frac{25 R}{8} \Rightarrow x = 25$

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

A monoatomic gas at pressure P and volume V is suddenly compressed to one eight of its original volume. The final pressure at constant entropy will be:

6 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Constant Entropy means Adiabatic process

$P v^{y} = C$

$P_{1} {v_{1}}^{y} = P_{2} {v_{2}}^{y}$

$P_{2} = P_{1} {(\frac{v_{1}}{v_{2}})}^{y} = P {(\frac{v}{\frac{1}{8} v})}^{y}$

$P_{2} = P (8) \frac{5}{3}$

P₂ = 32 P

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

The volume V of a given mass of monoatomic gas changes with temperature T according to the relation $V = k T^{\frac{2}{3}} .$  The work done when temperature changes by 90 K will be xR. The value of x is…….

[R = universal gas constant]

5 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

$V = k T^{\frac{2}{3}}$

$T V^{- \frac{3}{2}} = K$

$\Rightarrow γ - 1 = - 3 / 2$

$\Rightarrow γ = - 1 / 2$

Work done = $\frac{n R Δ T}{- y + 1} = \frac{1 \times R \times 90}{3 / 2} = 6 R$

n = 60

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R = 8.3 JK^-1 mol^-1)

6 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

$= 7 \times \frac{3}{2} \times 8.3 \times 40$

$= 3486 J$

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

The internal energy (U), pressure (P) and volume (V) of an ideal gas are related as U = 3PV + 4. The gas is

2 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

U = 3PV + 4

$\Rightarrow n C_{v} T = 3 P V + 4$

$\Rightarrow n \times \frac{f R T}{2} = 3 P V + 4$

$\Rightarrow f = 6 + \frac{8}{P V}$

$\Rightarrow f > 6 \Rightarrow p o l y a t o m i c$

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New answer posted•a year ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

Match the $C_{p} / C_{V}$ ratio for ideal gases with different type of molecules:

Match the correct option?

2 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Mono atomic $? C_{v} = \frac{3 R}{2} C_{p} = \frac{5 R}{2}$

Di- atomic $? C_{V} = \frac{5 R}{2} C_{p} = \frac{7 R}{2}$

(Rigid)

Di-atomic $? C_{v} = \frac{7 R}{2} C_{p} = \frac{9 R}{2}$

(Non-Rigid)

Tri-atomic $? C_{V} = 3 R C_{P} = 4 R$

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New answer posted•a year ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

Among the following the number of state variable is____________.

Internal energy (U)

Volume (V)

Heat (q)

Enthalpy (H)

1 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

State functions are

(a) Internal energy

(b) Volume

(c) Enthalpy

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New answer posted•a year ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

Dinitrogen and dioxygen, the main constituents of air do not react with each other in atmosphere to form oxides of nitrogen because

4 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$N_{2} + O_{2} \overset{V e r y h i g h T e m p}{?} 2 N O$

Reaction is endothermic and required very high temperature for comobination

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