Physics Thermodynamics

A container is divided into two chambers by a partition. The volume of first chamber is 4.5 litre and second chamber is 5.5 litre. The first chamber contain 3.0 moles of gas at pressure 2.0 atm and second chamber contains 4.0 moles of gas at pressure 3.0 atm. After the partition is removed and the mixture attains equilibrium, then, the common equilibrium pressure existing in the mixture is x * 10^-1 atm. Value of x is----------.

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Contributor-Level 10

$P = \frac{P_{1} V_{1} + P_{2} V_{2}}{V_{1} + V_{2}} = \frac{2 * 4.5 + 3 * 5.5}{10} = \frac{9 + 16.5}{10} = 25.5 * 1 0^{- 1} a t m$

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If one mole of an ideal gas at (P₁ V₁) is allowed to expand reversibly and isothermally (A to B) its pressure is reduced to one- half of the original pressure (see figure). This is followed by a constant volume cooling till its pressure is reduced to one-fourth of the initial value (B ® C). Then it is restored to its initial state by a reversible adiabatic compression (C to A). The net work done by the gas is equal to:

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Contributor-Level 10

n = 1 mole

$W_{A B} = n R T l n \frac{v_{2}}{v_{1}} = 1 * R T l n \frac{2 v_{1}}{v_{1}} = R T l n 2$

$W_{B C} = 0$

$W_{C A} = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = \frac{\frac{P_{1}}{4} * 2 V_{1} - P_{1} * V_{1}}{γ}$

$\Rightarrow W_{C A} = \frac{- R T}{2 (γ - 1)}$

$∴ W_{t o t a l} = R T [l n 2 - \frac{1}{2 (γ - 1)}]$

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In a certain thermodynamical process, the pressure of a gas depends on its volume as kV³. The work done when the temperature change from 100°C to 300°C will be……………. nR, where n denotes number of moles of a gas.

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Contributor-Level 10

P = kV³ Þ PV^-3 = k Þ Polytropic Process with index m = -3

W = $\frac{n R (T_{2} - T_{1})}{1 - m} = \frac{n R (300 - 100)}{1 - (- 3)} = 50 n R$ .

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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 :

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Contributor-Level 10

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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A reversible heat engine converts one-fourth of the heat input into work. When the temperature of the sink is reduced by 52K, its efficiency is doubled. The temperature is Kelvin of the source will be…………

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Contributor-Level 10

Efficiency in the Carnot engine is defined as :

$η = \frac{w o r k}{h e a t} = \frac{1}{4}$

$η = 1 - \frac{T_{c}}{T_{h}}, 2 η = 1 - \frac{T_{c} - 52}{T_{h}} \Rightarrow T_{h} = 208 K$

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Thermodynamic process is shown below on a P-V diagram for one mole of an ideal gas. If V₂ = 2V₁ then the ratio of temperature T₂/T₁ is

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Contributor-Level 10

Since equation of the gas process is given so we can convert it into T & V form.

$P V^{\frac{1}{2}} = C$

$\Rightarrow \frac{T}{\sqrt[]{V}} = C$

$\Rightarrow \frac{T_{1}}{V_{1}} = T_{2}$ $V_{2}$

$\Rightarrow \frac{T_{1}}{T_{2}} =$ = $\frac{1}{\sqrt[]{2}}$

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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)

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alok kumar singh

Contributor-Level 10

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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A monoatomic gas at pressure P and volume V is suddenly compressed to one eight of its original volume. The final pressure at constant entropy will be

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Contributor-Level 10

Constant Entropy means Adiabatic process

$P v^{y} = C$

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

$P_{2} = P_{1} {(\frac{v_{1}}{v_{2}})}^{y} = P {(\frac{v}{\frac{1}{8} v})}^{y}$

$P_{2} = P (8) \frac{5}{3}$

P2 = 32 P

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7 mol of a certain monoatomic ideal gas undergoes a temperature increase of 40K at constant pressure. The increase in the internal energy of the gas in this process is:

(Given R = 8.3 JK^-1 mol^-1)

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Contributor-Level 10

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

$= 7 * \frac{3}{2} * 8.3 * 40$

$= 3486 J$

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What will be the average value of energy for a monatomic gas in thermal equilibrium at temperature T?

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