Physics Thermodynamics

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New answer posted•a year ago

Physics Thermodynamics 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.

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|

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New answer posted•a year ago

Physics Thermodynamics Class 11th

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

6 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

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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New answer posted•a year ago

Physics Thermodynamics Class 11th

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 :

4 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

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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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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:

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

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

[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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New answer posted•a year ago

Physics Thermodynamics Class 11th

Match List I with List II.

List I                                                                                  List II

(a) Isothermal                                                                 (i) Pressure constant

(b) Isochoric                                                                    (ii) Temperature constant

(c) Adiabatic                                                                   (iii) Volume constant

(d) Isobaric                                                                      (iv) Heat content is constant

Choose the correct answer from the options given below:

11 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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]

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$Δ Q = Δ U + Δ W$

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