Thermodynamic Process

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New answer posted•a year ago

Thermodynamic Process Class 11th

A calorimeter of water equivalent $20 g$ <!-- [if gte msEquation 12]>20 g $180 g$ of water at $25^{?} C$ <!-- [if gte msEquation 12]>25?C $m$ <!-- [if gte msEquation 12]>m100?C $100^{?} C$ is mixed in it till the temperature of the mixture is $31^{?} C$ <!-- [if gte msEquation 12]>31?C

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by
Alok Kumar Singh | Contributor-Level 10

a year ago

Heat lost by steam $=$ Heat gained by water and calorimeter.

$\begin{array}{r} m \times 540 + m \times 1 \times (100 - 31) = 200 \\ 540 m + 69 m = 1200 \\ m = \frac{1200}{609} \approx 2 \end{array}$

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New answer posted•a year ago

Thermodynamic Process 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

Thermodynamic Process Class 11th

The entropy of any system is given by S = $α^{2} β l n [\frac{μ k R}{J β^{2}} + 3]$  Where a and b are the constants. $μ$ , J, k and R no. of moles, mechanical equivalent of heat, Bolzmann constant and gas constant respectively.  $[T a k e S = \frac{d Q}{T}]$

Choose the incorrect option from the following:

12 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by
Alok Kumar Singh | Contributor-Level 10

a year ago

N_a -> Avogadro’s Number

$J β^{2} \equiv [μ k R] \Rightarrow β^{2} \equiv [k R] \Rightarrow β \equiv [k] \equiv [M L^{2} T^{- 2} K^{- 1}]$

, and

-> $S \equiv α^{2} β \Rightarrow [M L^{2} T^{- 2} K^{- 1}] \equiv α^{2} [M L^{2} T^{- 2} K^{- 1}] \Rightarrow α \equiv [M^{0} L^{2} T^{0} K^{0}]$ Dimensionless