Thermodynamics

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

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Contributor-Level 10

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

and $T \propto \frac{1}{V^{γ - 1}}$

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

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a month ago

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?

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

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

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

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

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

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a month ago

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:

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Contributor-Level 10

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

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a month ago

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]

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Contributor-Level 10

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.

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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]

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Contributor-Level 10

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

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a month ago

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:

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Contributor-Level 10

-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

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a month ago

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.

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Contributor-Level 10

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|

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a month ago

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 :

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Contributor-Level 10

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

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a month ago

n mole of perfect gas undergoes a cyclic process ABCA (see figure) consisting of the following processes.

A->B : Isothermal expansion at temperature T so that the volume is doubled from V₁ to V₂ and Pressure changes from P₁ to P₂.

B->C : Isobaric compression at pressure P₂ to initial volume V₁.

C->A : Isobaric change leading to change of pressure from P₂ to P₁.

Total work done in the complete cycle ABCA is:

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Contributor-Level 10

$W_{A B} = n R T l n \frac{2 V_{1}}{V_{1}} = n R T l n 2 .$

$W_{B C} = P_{2} (V_{1} - 2 V_{1}) = - P_{2} V_{1} = - \frac{1}{2} n R T$

[At B, 2P₂ V₁ = nRT]

$W_{C A} = 0$ [CA is isochoric process].

$W_{A B C A} = W_{A B} + W_{B C} + W_{C A} = n R T (l n (2) - \frac{1}{2})$

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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]

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