Physics Thermodynamics

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

An ideal gas undergoes a thermodynamic process in which internal energy $(U)$ of the gas depends on pressure $(P)$ of the gas as $= a P^{4}$ , where a is a positive constant. Assuming gas to be monoatomic, the molar heat capacity of the gas for given process will be

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$U = a P^{4}$

$\frac{P V}{γ - 1} = a P^{4}$

$P^{- 3} V = a (γ - 1)$

$P V^{- 1 / 3} = c o n s t a n t$

$C o m p a r i n g w i t h P V^{x} = c o n s t a n t$

$x = - \frac{1}{3}$

$C = \frac{R}{γ - 1} + \frac{R}{1 - x} = \frac{R}{\frac{5}{3} - 1} + \frac{R}{1 + \frac{1}{3}} = \frac{3 R}{2} + \frac{3 R}{4}$

$C = \frac{9 R}{4}$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

A certain amount of gas is taken through a cyclic process (ABCDA) that has Two isobars, one isochore and one isothermal. The cycle can be represented on a P?V indicator diagram as :

30 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

P? V indicator diagram for isobaric

P? V indicator diagram for isochoric process

P? V indicator diagram for isothermal process

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New answer posted•a year ago

Physics Thermodynamics Thermodynamics

The temperature of equals masses of three different liquids x, y and z are 10°C, 20°C and 30°C respectively. The temperature of mixture when x is mixed with y is 16°C and that when y is mixed with z is 26°C. Temperature of mixture when x and z are mixed will be:

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

msx (16 – 10) = msy (20 – 16)

$\Rightarrow \frac{s_{x}}{s_{y}} = \frac{2}{3}$

$m s_{y} (26 - 20) = m s_{z} (30 - 26)$

$\Rightarrow \frac{s_{y}}{s_{z}} = \frac{2}{3}$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

Two carnot engines operate in series between a heat source at a temperature $T_{1} = 900 K$ and heat sink at temperature $T_{3} = 400 K$ . There is another reservoir at temperature $T_{2}$ , as shown, with $T_{1} > T_{2} > T_{3}$ . The two engines are equally efficient if $T_{2} =$ K

3 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

$η_{1} = 1 - \frac{T_{2}}{T_{1}}, η_{2} = 1 - \frac{T_{3}}{T_{2}}$

as $η_{1} = η_{2} \Rightarrow \frac{T_{2}}{T_{1}} = \frac{T_{3}}{T_{2}} \Rightarrow T_{2} = {(T_{1} T_{3})}^{\frac{1}{2}} = \sqrt[]{400 \times 900} = 600 K$

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New answer posted•a year ago

Physics Thermodynamics Class 11th

A hot body is being cooled in air according to Newton's law of cooling, the rate of fall of temperature being K times the difference of its temperature with respect to that of surroundings. The time after which the rate of heat loss of the body will be half of the maximum rate of heat loss is $\frac{1}{2 k} {l o g}_{e} ? n$ (The time is to be counted from the instant t = 0 ). The value of n is

2 Views|Posted a year ago

Asked by Raj Pandey

Answered by

Raj Pandey | Contributor-Level 9

a year ago

Newton's law of cooling, $\begin{array}{r} - \frac{d T}{d t} = k (T - T_{0}) \Rightarrow - \int \frac{d T}{T - T_{0}} = k \int d t \\ \Rightarrow {[- l n ? |T - T_{0}|]}_{T}^{(T + T_{0}) / 2} = k t (? Heat \propto (T - T_{0})) \\ \Rightarrow - l n ? |\frac{(T - T_{0}) / 2}{T - T_{0}}| = k t \Rightarrow - l n ? |\frac{1}{2}| = k t \Rightarrow t = \frac{l n ? 2}{k} = \frac{l n ? 4}{2 k} ∴ n = 4 \end{array}$