Physics Thermodynamics

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New answer posted•a year ago

Physics Thermodynamics 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

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New answer posted•a year ago

Physics Thermodynamics Class 11th

In the reported figure, hear energy absorbed by a system in going through a cyclic process is_______ pJ.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Heat absorbed in cyclic process = Work done = 100? Joule

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New answer posted•a year ago

Physics Thermodynamics Class 11th

The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is___________. (R is the universal gas constant)

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Since process is isochoric, so

$Q = n C_{v} Δ T = n (\frac{f}{2} R) Δ T = 4 (\frac{5}{2} \times R) (50) = 500 R$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

The temperature of 3.00 mol of an ideal diatomic gas in increased by 40.0°C without changing the pressure of the gas. The molecules in the gas rotate but do not oscillate. If the ratio of change in internal energy of the gas to the amount of work done by the gas is  $\frac{x}{10} .$ Then the value of x (round off to the nearest integer) is________.

(Given R = 8.31 J mol^-1K^-1)

12 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Since process is isochoric

So $Δ U = n C_{v} Δ T$

$Δ U = n (\frac{5}{2} R) Δ T - - (i) [C_{V} = \frac{5}{2} R]$

And external work

$Δ W = n R Δ T - - (i i)$

$\frac{5}{2} = \frac{x}{10} \Rightarrow x = 25.00$

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

A sample of gas with $γ = 1.5$  is taken through an adiabatic process in which the volume is compressed from 1200 cm³ to 300 cm³. If the initial pressure is 200 kPa. The absolute value of the work done by the gas in the process =_________J.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Adiabatic equation : $P_{1} V^{γ} = P_{2} V^{γ} \Rightarrow (200) {(1200)}^{γ} = P {(300)}^{γ}$

$\Rightarrow P = (200) {(4)}^{\frac{3}{2}} = 1600 K P a$

$W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = \frac{240 - 480}{1.5 - 1} = - 480 J$

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

A mixture of hydrogen and oxygen has volume 500 cm³, temperature 300K, pressure 400 kPa and mass 0.76g. The ratio of masses of oxygen to hydrogen will be:-

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Using Ideal gas Equation :

PV = nRT

$\Rightarrow n = \frac{P V}{R T} = \frac{400 \times 1 0^{3} \times 500 \times 1 0^{- 6}}{8.3 \times 100} = 0.008$

=> $m_{1} = 0.12, m_{2} = 0.64$

=> $\frac{m_{2}}{m_{1}} = \frac{16}{3}$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

For an ideal gas the instantaneous change in pressure ‘p’ with volume ‘v’ is given by the equation $\frac{d p}{d v} = - a p .$  If p = p₀ at v = 0 is the given boundary condition, then the maximum temperature one mole of gas can attain is:

(Here R is the gas constant)

10 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$\frac{d P}{d V} = - a P, a t V = V_{0}, P = P_{0}$

T_max =? for n = 1

$\frac{d P}{d V} = - a P$ (Given)

On integrating, we get

$\int_{P_{0}}^{P} \frac{d P}{P} = - a \int_{0}^{v} d V$

$T_{m a x} = \frac{P_{0}}{a e R} ? n = 1$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

A heat engine operates between a cold reservoir at temperature T₂ = 400K and a hot reservoir at temperature T₁. It takes 300J of heat from the hot reservoir and delivers 240J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be________K.

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Kindly go through the solution

$\frac{T_{1}}{400} = \frac{300}{240} \Rightarrow T_{1} = 500 K$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

The ratio of species heats $(\frac{C_{P}}{C_{V}})$ in terms of degree of freedom (f) is given by:

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$\frac{C_{P}}{C_{v}} = y = 1 + 2 / f$

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

An electric appliance supplies 6000 J/min heat to the system. If the system delivers a power of 90W. How long it would take to increase the internal energy by 2.5 × 10³ J?

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{Δ U}{t} = \frac{6000}{60} - 90$

= 10 J

$t = \frac{Δ U}{10} = \frac{2.5 \times 1 0^{3}}{10} = 2.5 \times 1 0^{2}$ second

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