Thermodynamics

The total kinetic energy associated with n moles of monoatomic gas is

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Contributor-Level 10

Total kinetic energy associated with 'n' moles of monoatomic gas is = $\frac{3}{2} n R T$

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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]

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

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

$\Rightarrow γ - 1 = - 3 / 2$

$\Rightarrow γ = - 1 / 2$

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

->n = 60

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

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U = 3PV + 4

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

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

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

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

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The ionization enthalpy of Na⁺ formation from Na_(g)is 495.8 kJ mol^-1, while the electron gain enthalpy of Br is - 325.0 kJ mol^-1. Given the lattice enthalpy of NaBr is - 728.4 kJ mol^-1. The energy for the formation of NaBr ionic solid is (-) -------------- * 10^-1kJ mol^-1.

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Kindly consider the following figure

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The reaction of cyanamide, NH₂CN_(s) with oxygen was run in a bomb calorimeter and $Δ U$ was found to be -742.24 kJ mol^-1. The magnitude of $Δ H_{298}$ for the reaction

$N H_{2} C N_{(S)} + \frac{3}{2} O_{2 (g)} \to N_{2 (g)} + O_{2 (g)} + H_{2} O_{(I)} i s$ ------------- kJ. (Rounded off to the nearest integer)

[Assume ideal gases and R = 8.314 J mol^-1 K^-1]

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Contributor-Level 10

$Δ U = - 742.24 k J / m o l e Δ H_{298} = ?$

NH₂CN (s) + $\frac{3}{2}$ O₂ (g) ® N₂ (g) + O₂ (g) + H₂O (l)

= $- 742.24 + \frac{1}{2} * \frac{8.314}{1000} * 298$

$- 742.24 + 1.238 = - 741.002 \approx - 741 k J / m o l$

So, magnitude of $Δ H$ = 741 kJ/mol.

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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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2.2g of nitrous oxide (N₂O) gas is cooled at a constant pressure of 1 atm from 310 K to 270 K causing the compression of the gas from 217.1mL to 167.75mL. The change in thermal energy of the process, $Δ U$ is '-x' J. The value of 'x' is________. [Nearest Integer]

(Given : atomic mass of N = 14 g mol^-1 and of O = 16 g mol^-1.

Molar heat capacity of N₂O is 100 J K^-1 mol^-1)

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Contributor-Level 10

Moles of N₂O= $\frac{2.2}{44} = \frac{1}{20}$

$Δ H = n C_{p} Δ T = \frac{1}{20} * 100 (- 40) = - 200 J$

$Δ U = q_{p} + w$

$w = - P_{e x t} . Δ V$

$w = - 1 \frac{(167.75 - 217.1)}{1000} * 101.3 J = + 5 J$

$Δ U = - 200 + 5 = 195 J$

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Thermodynamics Class 12th

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

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Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

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Process-AB $\Rightarrow$ Isobaric,

Process-AC $\Rightarrow$ Isothermal, and

Process-ADÞAdiabatic

W₂ < W₁ < W₃

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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

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