Physics Thermodynamics

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New answer posted•a year ago

Physics Thermodynamics Class 11th

A thermodynamic system is taken from an original state D to an intermediate sate E by the linear pressure shown in the figure. Its volume is then reduced to the original volume from E to F by an isobaric process. The total work done by the gas from D to E to F will be

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

W_total = W_D→E + W_E→F = Area of triangle DEF

= $\frac{1}{2} \times 3 \times 300 = 450 J .$

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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 Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

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

In a carnot engine, the temperature of reservoir is 527°C and that of sink is 200 K. If the work done by the engine when it transfers heat from reservoir to sink is 12000 kJ, the quantity of heat absorbed by the engine from reservoir is___________× 10⁶ J.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

T₂ = 200 k

$T_{1} = 527 ° C + 273 = 800 K$

w = 12000 kJ = 12 × 10⁶ J

Q₁ =?

$x = 1 - \frac{T_{2}}{T_{1}} = \frac{w}{Q_{1}} = 1 - \frac{200}{800} = \frac{12 \times 1 0^{6}}{Q_{1}}$

$\Rightarrow \frac{3}{4} = \frac{12 \times 1 0^{6}}{Q_{1}}$

$\Rightarrow Q_{1} = 16 \times 1 0^{6} J$

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

A mixture of hydrogen and oxygen has volume 2000 cm³, temperature 300K, pressure 100 kPa and mass 0.76g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:

[Take gas constant R = 8.3 JK^-¹ mol^-¹]

24 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$V_{T} = 2000 c m^{3} = 2000 \times 1 0^{- 6} m^{3} (T o t a l v o l u m e o f m i x t u r e)$

T = 300 K

$_{}^{}^{}$ $P_{T} = 100 k P a = 100 \times 1 0^{3} P a = 1 0^{5}$ Pa

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

$P_{T} V = (x_{1} + x_{2}) R T$

$\Rightarrow 1 0^{5} \times 200 \times 1 0^{- 6} = (x_{1} + x_{2}) \times 8.3 \times 300$

$\frac{x_{1}}{x_{2}} = \frac{0.06}{0.02} = 3 : 1$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

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?

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

A certain amount of gas of volume V at $27 ° C$  temperature and pressure $2 \times 1 0^{7} N m^{- 2}$  expands isothermally until its volume gets doubled. Late it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be

(Use $γ$ = 1.5):

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$P_{1} = 2 \times 1 0^{7} P a$

$P_{1} v_{1} = P_{2} v_{2}$

Since v₂ = 2v₁ Hence $P_{2} = \frac{P_{1}}{2} (I s o t h e r m a l E x p a n s i o n)$

$P_{2} = 1 \times 1 0^{7} P a$

$P_{2} {(v_{2})}^{y} = P_{3} {(2 v_{2})}^{y}$

$P_{3} = \frac{1 \times 1 0^{7}}{2^{1.5}} = 3.536 \times 1 0^{6} P a$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

Let $η_{1}$  is the efficiency of an engine at T₁ = $447 ° C$  and T₂ = $147 ° C$  while $η_{2}$  is the efficiency at T₁ = $947 ° C$  and T₂ $47 ° C$ . The ratio $\frac{η_{1}}{η_{2}}$  will be

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

As we know that

$η = 1 - \frac{T_{2}}{T_{1}}, s o$

$\frac{n_{1}}{n_{2}} = \frac{0.42}{0.74} \approx 0.56$

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

A Carnot engine operates between two reservoirs of temperatures $900 K$ and $300 K$ . The engine performs $1200 J$ of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$n = \frac{W}{Q_{h}} = 1 - \frac{300}{900} = \frac{2}{3}$

$Q_{h} = \frac{3}{2} W = 1800 J$

$Q_{L} = O_{h} - W = 600 J$

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New question posted•a year ago

Physics Thermodynamics Thermodynamics

A Carnot engine operates between two reservoirs of temperatures   and  . The engine performs   of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is

3 Views|Posted a year ago

Asked by Vishal Baghel

New answer posted•a year ago

Physics Thermodynamics Class 11th

A Block of ice of mass 120g at temperature $0 ° C$  is put in 300g of water at $25 ° C .$  The xg of ice melts as the temperature of the water reaches $0 ° C .$  The value of x is _________.

[Use specific heat capacity of water = 4200Jkg^-1K^-1, Latent heat of ice = 3.5 × 10⁵Jkg^-1]

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Using concept of calorimeter, we can write

Gain in heat = Loss in heat

$x \times L = m c Δ T \Rightarrow x \times 3.5 \times 1 0^{5} = \frac{300}{1000} \times 25 \times 4200$

$\Rightarrow x = \frac{300 \times 25 \times 4200}{1000 \times 3.5 \times 1 0^{5}} = \frac{3 \times 25 \times 42}{35 \times 1000} = \frac{90}{1000} k g = 90 g m$

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