Physics Thermodynamics

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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4 months ago

For a gas C? - C? = R in a state P and C? - C? = 1.10 R in a state Q, T? and T? are the temperatures in two different states P and Q respectively. Then

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Raj Pandey

Contributor-Level 9

C? = dQ/ndT = (dU + pdV)/ndT
= C? + (pdV/ndT)
C? - C? = pdV/ndT = R - For ideal Gas [Box: PV = nRT, pdV = nRdT]
C? - C? = 1.1R - For Non – Idea gas (for Real gas)
And Real gas behaves as ideal gas at high temperature & low pressure.
∴ T_B > T_A

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4 months ago

A monoatomic ideal gas, initially at temperature T?, is enclosed in a cylinder fitted with a frictionless piston. The gas is allowed to expand adiabatically to a temperature T? by releasing the piston suddenly. If ? and ? are the lengths of the gas column, before and after the expansion respectively, then the value of T?/T? will be :

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Raj Pandey

Contributor-Level 9

For adiabatic process
T? V? ¹ = T? V? ¹
⇒ T? (Al? )? ¹ = T? (Al? )? ¹
T? /T? = (l? /l? )? ¹ = (l? /l? )? /³? ¹ = (l? /l? )?

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4 months 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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alok kumar singh

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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4 months 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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alok kumar singh

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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4 months ago

Match List I with List II.

List I List II

(a) Isothermal &nb

...more

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alok kumar singh

Contributor-Level 10

In isothermal process, temperature is constant.

In isochoric process, volume is constant.

In adiabatic process, there is no exchange of heat.

In isobaric process, pressure is constant

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4 months ago

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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Contributor-Level 10

$Δ Q = Δ U + Δ W$

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

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

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4 months ago

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]

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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

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4 months ago

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

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Contributor-Level 10

U = 3PV + 4

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

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

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

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

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4 months ago