Thermodynamics

Get insights from 325 questions on Thermodynamics, answered by students, alumni, and experts. You may also ask and answer any question you like about Thermodynamics

325
Questions
0
Discussions
0
Active Users
0
Followers

Have a question about Thermodynamics? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Thermodynamics Class 11th

Two moles of an ideal gas with $\frac{C_{P}}{C_{V}} = \frac{5}{3}$ $\frac{_{}}{_{}} \frac{}{}$ are mixed with 3 moles of another ideal gas with $\frac{C_{P}}{C_{V}} = \frac{4}{3}$ $\frac{_{}}{_{}} \frac{}{}$ . The value of $\frac{C_{P}}{C_{V}}$ $\frac{_{}}{_{}}$ for the mixture is:

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$γ_{mixture} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} C_{V_{1}} + n_{2} C_{V_{2}}} = \frac{n_{1} \frac{γ_{1} R}{γ_{1} - 1} + n_{2} \frac{γ_{2} R}{γ_{2} - 1}}{\frac{n_{1} R}{γ_{1} - 1} + \frac{n_{2} R}{γ_{2} - 1}}$ $_{} \frac{_{}_{_{}}_{}_{_{}}}{_{}_{_{}}_{}_{_{}}} \frac{_{} \frac{_{}}{_{}}_{} \frac{_{}}{_{}}}{\frac{_{}}{_{}} \frac{_{}}{_{}}}$

on rearranging we get

$\frac{n_{1} + n_{2}}{γ_{mix} - 1} = \frac{n_{1}}{γ_{1} - 1} + \frac{n_{2}}{γ_{2} - 1}; \frac{5}{γ_{mix} - 1} = \frac{3}{1 / 3} + \frac{2}{2 / 3}$

$\frac{5}{γ_{mix} - 1} = 9 + 3 = 12 \Rightarrow γ_{mixture} = \frac{17}{12} + 1 + \frac{5}{12}; γ_{mix} = 1.42$

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

Under an adiabatic process, the volume of an ideal gas gets doubled. Consequently the mean collision time between the gas molecule changes from $τ_{1}$ to $τ_{2}$ . If $\frac{C_{p}}{C_{v}} = γ$ for the gas then a good estimate for $\frac{τ_{2}}{τ_{1}}$ is given by

15 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Relaxation time $(τ) \propto \frac{V}{\sqrt[]{T}}$

and $T \propto \frac{1}{V^{γ - 1}}$

$\begin{array}{r} τ \propto V^{1 + \frac{γ - 1}{2}} τ \propto V^{\frac{1 + γ}{2}} \\ \frac{τ_{f}}{τ_{i}} = {(\frac{2 V}{V})}^{\frac{1 + γ}{2}} \frac{τ_{f}}{τ_{i}} = (2)^{\frac{1 + γ}{2}} \end{array}$

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

Two ideal Carnot engines operate in cascade (all heat given up by one engine is used by the other engine to produce work) between temperatures, $T_{1}$ and $T_{2}$ . The temperature of the hot reservoir of the first engine is $T_{1}$ and the temperature of the cold reservoir of the second engine is $T_{2}$ . $T$ is temperature of the sink of first engine which is also the source for the second engine. How is $T$ related to $T_{1}$ and $T_{2}$ , if both the engines perform equal amount of work?

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$W = Q_{1} - Q_{2} = Q_{2} - Q_{3}$

$Q_{2} = \frac{Q_{1} + Q_{3}}{2}$

$2 = \frac{Q_{1}}{Q_{2}} + \frac{Q_{3}}{Q_{2}}$

$2 = \frac{T_{1}}{T} + \frac{T_{2}}{T}$

$T = \frac{T_{1} + T_{2}}{2}$

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

Two Carnot engines A and B operate in series such that engine A absorbs heat at T? and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at T?. When work done in both the cases is equal, the value of T is:

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

η_A = W_A/Q? = 1 - T/T?
η_B = W_B/Q? ' = 1 - T? /T
W_A = Q? (1 - T/T? ) = Q? - Q?
W_B = Q? ' (1 - T? /T) = Q? /2 (1 - T? /T)
Given W_A = W_B
Q? (1 - T/T? ) = (Q? /2) (1 - T? /T)
Q? (T? /T) (1-T/T? ) = (Q? /2) (1-T? /T)
(T? /T - 1) = (1/2) (1-T? /T)
2T? /T - 2 = 1 - T? /T
2T? /T + T? /T = 3
T = (2T? +T? )/3

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

One mole of an ideal gas is taken through an adiabatic process where the temperature rises from 27°C to 37°C. If the ideal gas is composed of polyatomic molecule that has 4 vibrational modes, which of the following is true? [R = 8.314 J mol⁻¹K⁻¹]

11 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Degrees of freedom (f) = 3 translational + 3 rotational + (2 × 4) vibrational = 14
γ = 1 + 2/f = 1 + 2/14 = 8/7
W = nR? T/ (γ-1) = (1 × 8.3 × (310-300)/ (8/7 - 1) = (83)/ (1/7) = 581 J
As W is positive, work is done by the gas. The solution says W<0, work done on the gas. This implies? T is negative. The question states temperature rises, so work is done on the gas. W = -582J.

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

For water at 100°C and 1 bar, ΔᵥₐₚH - ΔᵥₐₚU = _______ × 10² J mol⁻¹ (Round off to the Nearest integer) [Use: R = 8.31 J mol⁻¹ K⁻¹] [Assume volume of H₂O(l) is much smaller than volume of H₂O(g). Assume H₂O(g) can be treated as an ideal gas]

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

H? O (l) → H? O (g)
ΔH° = ΔU° + ΔngRT
ΔH° - ΔU° = ΔngRT
= 1 × 8.31 × 373
= 3099.63 J/mol
= 30.9963 × 10² J/mol
≈ 31 × 10² J/mol

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

A body takes 4 min. to cool from 61°C to 59°C. If the temperature of the surroundings is 30°C, the time taken by the body to cool from 51°C to 49°C is:

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

-dT/dt = K [T - Ts]

(61-59)/4 = K [ (61+59)/2 - 30]
-0.5 = K [60 - 30] = 30K
So, K = -1/60 min? ¹
Again
(51-49)/t = K [ (51+49)/2 - 30]
-2/t = (-1/60) [50-30] = -20/60 = -1/3
t = 6 min

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

In the reported figure, there is a cyclic process ABCDA on a sample of 1mol of a diatomic gas. The temperature of the gas during the process A→B and C→D are T? and T? (T? > T?) respectively. Choose the correct option out of the following for work done if processes BC and DA are adiabatic.

12 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

A→B & D→C (isothermal process)
So, TA = TB & TD = TC. Now B→C & D→A (adiabatic process)
|WBC| = nR/ (γ-1) (TB - TC)
|WAD| = nR/ (γ-1) (TA - TD) = nR/ (γ-1) (TB - TC)
∴ |WBC| = |WAD|

Upvote

New answer posted•a year ago

Thermodynamics Class 11th

A heat engine has an efficiency of 1/6. When the temperature of sink is reduced by 62°C, its efficiency get doubled. The temperature of the source is :

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

η = 1 - T? /T?
1/6 = 1 - T? /T? (i)
2η = 1/3 = 1 - (T? -62)/T? (ii)
By (i) and (ii)
(T? /T? ) = 5/6
1/3 = 1 - (T? /T? ) + 62/T? = 1 - 5/6 + 62/T?
1/3 = 1/6 + 62/T?
1/6 = 62/T?
T? = 62 × 6 = 372K = 99°C

Upvote

New answer posted•a year ago

Thermodynamics Class 11th