Physics Thermodynamics

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 :

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P? V indicator diagram for isobaric

P? V indicator diagram for isochoric process

P? V indicator diagram for isothermal process

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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:

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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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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

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

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

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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

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Contributor-Level 9

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

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12gm of gas occupy a volume of 4 * 10^-3 m³ at a temperature of 7°C. After the gas is heated at constant pressure its density becomes 6 * 10^-4gm/cm³. What is the temperature to which the gas was heated (in kelvin)?

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P = (RT/M? )ρ ⇒ P/ρT = constant
⇒ (12)/ (4 * 10? ³ * 10? ) * 280 = 6 * 10? * T ⇒ T = 1400K

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The molar specific heat C_v of a diatomic gas is (5/2)R when we treat its molecules to be rigid having no vibrations. But this assumption is not always valid especially at high temperatures. In that case, molecules are treated as non-rigid having vibrational energy also. C_v for diatomic gas with non-rigid gas molecules will be:

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Vibrational energy of a non-rigid gas molecule is K? T so, total energy = (5/2)K? T + K? T = (7/2)K? T
∴ C? = (7/2)R

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Steam at $100^{?} C$ is passed into a calorimeter of water equivalent 10 gm containing 74gm of water and 10gm of ice at . If the temperature of the calorimeter and its contents rises to $0^{?} C$ , the amount of steam passed (Latent heat of vapourisation $= 540 c a l / g$ , latent heat of fusion $= 80 c a l / g$ ) is:

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Heat lost $=$ = Heat gained

$\Rightarrow m * 540 + m * s_{ω} * (100 - 5) = 10 * 80 + (10 + 74 + 10) * s_{ω} * 5$

$\Rightarrow m (540 + 95) = 800 + 94 * 5$

$\Rightarrow m 635 = 1270$

$∴ m = 2 g m$

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Three samples of the same gas A,B and $C (γ = 3 / 2)$ have initially equal volume. Now the volume of each sample is doubled. The process is adiabatic for A, isobaric for B and isothermal for C. If the final pressures are equal for all three samples, the ratio of their initial pressures are:

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For gas $A : P_{A} ($ $_{}$ initial $) = {(\frac{v_{f}}{v_{i}})}^{γ} P = (2)^{3 / 2} P$ ${(\frac{_{}}{_{}})}^{}^{}$

$P_{A} = 2 \sqrt[]{2} P$

For gas B: $P_{B} ($ $_{}$ initial $) = P$

For gas $C : P_{C} ($ initial $) = \frac{2 V_{0}}{V_{0}} P = 2 P ∴$ Ratio $= 2 \sqrt[]{2} : 1 : 2$

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The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is:

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alok kumar singh

Contributor-Level 10

$η = \frac{W . D . b y g a s}{H e a t a b s o r b}$

$η = \frac{H e a t (N e t)}{H e a t (A b s o r b)}$

$η = \frac{A r e a u n d e r t h e c u r v e}{H e a t A b s o r b}$

$η = \frac{\frac{1}{2} * S_{0} T_{0}}{\frac{1}{2} (3 T_{0}) S_{0}} = \frac{1}{3}$

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A carnot engine works between temperatures 327°C and 27°C. If the engine takes 1600 J of heat from the higher temperature reservoir, the work done by the engine per cycle (in Joule) is equal to ______.

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