Thermodynamics

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New answer posted•a year ago

Thermodynamics Class 11th

A mixture of hydrogen and oxygen has volume 2000 cm³, temperature 300K, pressure 100 kPa and mass 0.76g. The ratio of number of moles of hydrogen to number of moles of oxygen in the mixture will be:

[Take gas constant R = 8.3 JK^-¹ mol^-¹]

24 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$V_{T} = 2000 c m^{3} = 2000 \times 1 0^{- 6} m^{3} (T o t a l v o l u m e o f m i x t u r e)$

T = 300 K

$_{}^{}^{}$ $P_{T} = 100 k P a = 100 \times 1 0^{3} P a = 1 0^{5}$ Pa

mT = 8.3 J/K1 mol1

Now, using Ideal gas equation

$P_{T} V = (x_{1} + x_{2}) R T$

$\Rightarrow 1 0^{5} \times 200 \times 1 0^{- 6} = (x_{1} + x_{2}) \times 8.3 \times 300$

$\frac{x_{1}}{x_{2}} = \frac{0.06}{0.02} = 3 : 1$

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New answer posted•a year ago

Thermodynamics Class 11th

A certain amount of gas of volume V at $27 ° C$  temperature and pressure $2 \times 1 0^{7} N m^{- 2}$  expands isothermally until its volume gets doubled. Late it expands adiabatically until its volume gets redoubled. The final pressure of the gas will be

(Use $γ$ = 1.5):

4 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$P_{1} = 2 \times 1 0^{7} P a$

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

Since v₂ = 2v₁ Hence $P_{2} = \frac{P_{1}}{2} (I s o t h e r m a l E x p a n s i o n)$

$P_{2} = 1 \times 1 0^{7} P a$

$P_{2} {(v_{2})}^{y} = P_{3} {(2 v_{2})}^{y}$

$P_{3} = \frac{1 \times 1 0^{7}}{2^{1.5}} = 3.536 \times 1 0^{6} P a$

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New answer posted•a year ago

Thermodynamics Class 12th

A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be:

8 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Efficiency $η = 50 %$

$\Rightarrow η = (1 - \frac{T_{2}}{T_{1}}) \times 100 = 50$

$\Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{50}{100} \Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{1}{2}$ -(i)

After the change :- $1 - \frac{T_{2}^{1}}{T_{1}} = \frac{50 \times (1.3)}{100} = \frac{65}{100} = . 65$ $\frac{_{}^{}}{_{}} \frac{}{} \frac{}{}$ -(ii)

$\Rightarrow 1 - (\frac{T_{2} - 40}{T_{1}}) = . 65$ --(ii)

$\Rightarrow \frac{40}{T_{1}} = . 65 - . 50 = . 15 \Rightarrow T_{1} = \frac{4000}{15} = 266.7 k$

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New answer posted•a year ago

Thermodynamics Class 11th

A Carnot engine operates between two reservoirs of temperatures $900 K$ and $300 K$ . The engine performs $1200 J$ of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is

7 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$n = \frac{W}{Q_{h}} = 1 - \frac{300}{900} = \frac{2}{3}$

$Q_{h} = \frac{3}{2} W = 1800 J$

$Q_{L} = O_{h} - W = 600 J$

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New question posted•a year ago

Thermodynamics Class 11th

<p><strong>A Carnot engine operates between two reservoirs of temperatures   and  . The engine performs   of work per cycle. The heat energy (in J) delivered by the engine to the low temperature reservoir, in a cycle is</strong></p>

3 Views|Posted a year ago

Asked by Vishal Baghel

New answer posted•a year ago

Thermodynamics Class 11th

A gas (Molar mass = 280 g mol^-¹) was burnt in excess O₂ in a constant volume calorimeter and during combustion the temperature of calorimeter increased from 298.0 K to 298.45K. If the heat capacity of calorimeter is 2.5 kJ K^-¹ and enthalpy of combustion of gas is 9 kJ mol^-¹ the amount of gas burnt is _____________g. (Nearest Integer)

19 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Let x gm is burnt

Moles = x/280

Heat released by x/280 mol = 2.5 × 0.45 kJ

Heat released by 1 mol = $\frac{2.5 \times 0.45 \times 280}{x} = k J$

$\begin{array}{l} Δ H = Δ U + Δ n g R T \\ Δ H = Δ U \end{array}$

$9 = \frac{2.5 \times 280 \times 0.45}{x}$

x = 35 gm

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New answer posted•a year ago

Thermodynamics Class 11th

The enthalpy of combustion of propane, graphite and dihydrogen at 298 K are -2220.0 kJ mol^-1, -393.5kJ mol^-1 and -285.8kJ mol^-1 respectively. The magnitude of the enthalpy of formation of propane C₃H₈ is _________ kJ mol^-1 (Nearest integer)

7 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

3C (Graphite) + 4H₂ (g) -> C₃H₈ (g)

$Δ H_{f} (C_{3} H_{8}) = [3 \times Δ H_{c o m b} (C)] + [4 \times Δ H_{c o m b} (H_{2})] - [Δ H_{c o m b} (C_{3} H_{8})]$

= -10.3.7 kJ/mole

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New answer posted•a year ago

Thermodynamics Class 11th

The pressure P₁ and density d₁ of diatomic gas $(γ = \frac{7}{5})$  changes suddenly to P₂(>P₁) and d₂ respectively during an adiabatic process. The temperature of the gas increases and becomes ________times of its initial temperature. (given  $\frac{d_{2}}{d_{1}} = 32)$

2 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

For adiabatic process – PV^Y= const

$T_{1} {V_{1}}^{Y - 1} = T_{2} V_{2}^{Y - 1}$

$\frac{T_{2}}{T_{1}} = {(\frac{v_{1}}{v_{2}})}^{Y - 1} = {(\frac{d_{2}}{d_{1}})}^{Y - 1} = {(32)}^{(\frac{7}{5} - 1)} = {(32)}^{2 / 5}$

$= {(2)}^{2} = 4$

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New answer posted•a year ago

Thermodynamics Class 11th

2O₃(g) $?$ 3O₂(g)

At 300 K, ozone is fifty percent dissociated. The standard free energy change at this temperature and 1 atm pressure is (-)________. J mol^?1. (Nearest integer)

[Given: In 1.35 = 0.3 and R=8.3 J K^-1 mol^-1]

12 Views|Posted a year ago

Asked by Vishal Baghel

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Answered by

Vishal Baghel | Contributor-Level 10

a year ago

2O₃(g) -> 3O₂

t = 0 a moles 0

-0.5a mole +0.75a mole

At Eq. 0.5a mole 0.75a mole

Total moles at eq = 0.5a + 0.

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New answer posted•a year ago

Thermodynamics Class 11th

A carnot engine takes 5000 kcal of heat from a reservoir at $727 ° C$  and gives heat to a sink at $127 ° C .$  The work done by the engine is

2 Views|Posted a year ago

Asked by Payal Gupta

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Answered by

Payal Gupta | Contributor-Level 10

a year ago

T₁ = 727 + 273 = 1000k

T₂ = 127° + 273 = 400 k

Q₁ = 5 × 10³ k cal

$\Rightarrow \frac{600}{1000} = \frac{w}{5000}$

w = 12.6 × 10⁶ J

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