Physics Thermodynamics

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New answer posted•a year ago

Physics Thermodynamics Class 11th

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  The value of x is $\frac{x R}{8} .$ ………

[R = universal gas constant]

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$Δ Q = Δ U + Δ W$

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

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

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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]

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$V = k T^{\frac{2}{3}}$

-> $T V^{- \frac{3}{2}} = K$

$\Rightarrow γ - 1 = - 3 / 2$

$\Rightarrow γ = - 1 / 2$

Work done = $\frac{n R Δ T}{- y + 1} = \frac{1 \times R \times 90}{3 / 2} = 6 R$

->n = 60

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

U = 3PV + 4

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

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

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

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

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

A diatomic gas, having C_P $= \frac{7}{2} R a n d C_{V} = \frac{5}{2} R,$ is heated at constant pressure. The ratio dU : dQ : dW :

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

At constant pressure,

$d U = n C_{V} d T = n . \frac{5}{2} R d T$

$d Q = n C_{P} d T = n . \frac{7}{2} R d T$

dW = nRdT

dU : dQ : dW = 5 : 7 : 2

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New answer posted•a year ago

Physics Thermodynamics Class 11th

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed  and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by:

(R = universal gas constant)

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$K E = Δ U$

$Δ U = \frac{1}{2} m v^{2}$

$n C_{v} Δ T = \frac{m v^{2}}{2}$

$\Rightarrow \frac{m}{M} \frac{R}{y - 1} Δ T = \frac{m v^{2}}{2}$

$Δ T = \frac{0.4}{2} \frac{M v^{2}}{R}$

$Δ T = \frac{M v^{2}}{5 R}$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

The efficiency of a Carnot’s engine, working between stem point and ice point, will be:

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

Steam point (T₁) = 100°C = 273 + 100 = 373 K

Ice point (T₂) = 0°C = 273 K

$η = (1 - \frac{T_{2}}{T_{1}}) \times 100 = (1 - \frac{273}{373}) \times 100 = 26.81 %$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

The total internal energy of two mole mono atomic ideal gas at temperature T = 300K will be __________J. (Given R = 8.31 J/mol.K)

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$U = n c_{v} T$

$\begin{array}{l} = 2 \times \frac{3}{2} \times R \times 300 \\ = 7479 J \end{array}$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

Given below are two statements:

Statement – I : When $μ$ moment of an ideal gas undergoes adiabatic change from state $(P_{1}, V_{1}, T_{1})$ to state $(P_{2}, V_{2}, T_{2})$ then work done is w = $\frac{μ R (T_{2} - T_{1})}{1 - γ},$ where $γ = \frac{C_{p}}{C_{v}} a n d$ R = universal gas constant.

Statement – II : In the above case, when work done on the gas, the temperature of the gas would rise.

Choose the correct answer from the options given below:

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

In case of adiabatic process -

$W o r k d o n e, W = \frac{η R (T_{2} - T_{1})}{1 - Y}$

$A s, Δ Q = Δ U + W$

for adiabatic process :- $Δ Q = 0$

$\Rightarrow Δ U = - W$

So, when work is done by the gas, temperature decreases and when work is done on the gas, temperature rises.

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

Starting with the same initial conditions, an ideal gas expands form volume V₁ to V₂ in the three different ways. The work done by the gas in W₁ if the process is purely isothermal, W₂, if the process is purely adiabatic and W₃ if the process is purely isobaric. Then, choose the correct option

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

Process-AB $\Rightarrow$ Isobaric, &n

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New answer posted•a year ago

Physics Thermodynamics Class 11th

In an experiment to verigy Newton’s law of cooling, a graph is plotted between, the temperature difference $(Δ T)$ of the water and surrounding and time as shown in figure. The initial temperature of water is taken as $80 ° C .$ The value of t₂ as mentioned in the graph will be _______.

12 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$\frac{d T}{d t} = - k (T - T_{0})$

$\frac{d T}{(T - T_{0})} = - k d t$

$\Rightarrow k = - \frac{1}{6} l n (\frac{2}{3})$ - (1)

Now $\frac{d T}{d t} = - k (T - T_{0})$

t = 10.257

t₂ = 6 + 10.257 = 16.257 minutes

16.25 and or 16

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