Thermodynamics

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New answer posted•a year ago

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

[R = universal gas constant]

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | 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

Thermodynamics Class 11th

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]

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | 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

Thermodynamics Class 11th

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

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | 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

Thermodynamics Class 11th

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:

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

n = 1 mole

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

$W_{B C} = 0$

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

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

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

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New answer posted•a year ago

Thermodynamics Class 11th

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.

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Thermodynamics Class 11th

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 :

3 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

Thermodynamics Class 11th

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)

33 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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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New answer posted•a year ago

Thermodynamics Class 12th

In 5 minutes, a body cools from 75°C to 65°C at room temperature of 25°C. The temperature of body at the end of next 5 minutes is__________°C.

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

from newtan’s law of cooling :

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

Using average value :

$\frac{75 - 65}{5} = - k [\frac{(75 + 65) 2}{2} - 25]$

$\frac{(65 - T)}{5} = - k [\frac{(65 + T)}{2} - 25]$

$\Rightarrow \frac{10}{65 - T} = \frac{45}{\frac{65 + T}{2} - 25}$

$\Rightarrow T = 57 ° C$

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New answer posted•a year ago

Thermodynamics Class 11th

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

3 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Total energy of monoatomic gas at equilibrium

$= \frac{3}{2} K_{B} T$

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New answer posted•a year ago

Thermodynamics Class 11th

For a given chemical reaction A ® B at 300 K at the free energy change is – 49.4 kJ mol^-1 and the enthalpy of reaction is 51.4 kJ mol^-1. The entropy change of the reaction is________J K ^-1 mol^-1_.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

From $Δ H = Δ G + T Δ S ∴ Δ S = \frac{Δ H - Δ G}{T}$

$∴$ $Δ S = [\frac{51.4 - (- 49.4)}{300}] \times 1000 J K^{- 1} m o l^{- 1} = 336 J K^{- 1} m o l^{- 1}$

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