Tags Thermodynamics

Thermodynamics

Get insights from 324 questions on Thermodynamics, answered by students, alumni, and experts. You may also ask and answer any question you like about Thermodynamics

Follow Ask Question

324

Questions

Discussions

Active Users

Followers

New answer posted

2 months 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)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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)

= $2 * \frac{R}{γ - 1} Δ T$

= $\frac{2 * R}{\frac{5}{3} - 1} * 20$

= $= 2 * \frac{3 R}{2} * 20$

Q = 60R = 60 * 8.3

Q = 498 J

0 0

New answer posted

2 months 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:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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

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

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

0 0

New answer posted

2 months 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]

0 Follower 4 Views

Payal Gupta

Contributor-Level 10

$Δ Q = Δ U + Δ W$

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

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

0 0

New answer posted

2 months 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:

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

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

0 0

New answer posted

2 months 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]

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

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

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

$\Rightarrow γ - 1 = - 3 / 2$

$\Rightarrow γ = - 1 / 2$

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

n = 60

0 0

New answer posted

2 months 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)

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

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

$= 3486 J$

0 0

New answer posted

2 months 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

0 Follower 2 Views

Payal Gupta

Contributor-Level 10

U = 3PV + 4

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

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

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

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

0 0

New answer posted

2 months 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?

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

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 $? C_{V} = 3 R C_{P} = 4 R$

0 0

New answer posted

2 months 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)

0 Follower 1 View

Vishal Baghel

Contributor-Level 10

State functions are

(a) Internal energy

(b) Volume

0 0

New answer posted

2 months 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

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

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

Reaction is endothermic and required very high temperature for comobination

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Thermodynamics

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience