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3 months ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Six

The change in the magnitude of the volume of an ideal gas when a small additional pressure $Δ P$ is applied at a constant temperature, is the same as the change when the temperature is reduced by a small quantity $Δ T$ at constant pressure. The initial temperature and pressure of the gas were $300 K$ and 2 atm. respectively. If $| Δ T | = C | Δ P |$ then value of $C$ in (K/atm) is .

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alok kumar singh

Contributor-Level 10

Sol.

$\begin{array}{l} A H \to \frac{H}{2} & H = \frac{\sqrt[]{3} a}{2} \\ A G \to \frac{H}{\sqrt[]{3}} & A l = \frac{H}{2 \sqrt[]{3}} \\ M_{A B C} = M & M_{A D E} = \frac{M}{4} \end{array}$

$\Rightarrow I_{G} = \frac{M a^{2}}{12} - [\frac{M}{4} \frac{{[\frac{a}{2}]}^{2}}{12} + \frac{M}{4} {[\frac{a}{2 \sqrt[]{3}}]}^{2}] = (\frac{11}{16}) \frac{{M a}^{2}}{12}$

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3 months ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Six

Match the thermodynamics processes taking place in a system with the correct conditions. In the table: $Δ Q$ is the heat supplied, $Δ W$ is the work done and $Δ U$ is change in internal energy of the system. Match the following:

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alok kumar singh

Contributor-Level 10

Sol. $? v$

$\Rightarrow m v = 16 {m v}_{1}$

$\Rightarrow V_{1} = \frac{V}{16}$

$\Rightarrow Δ k$ loss $= \frac{1}{2} {m v}^{2} - \frac{1}{2} (16 m) {(\frac{V}{16})}^{2}$

$\Rightarrow \frac{1}{2} {m v}^{2} (\frac{15}{16})$

$%$ loss $= \frac{15}{16} * 100 = 93.75 %$

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3 months ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 12th Chapter Six

When 5 moles of He gas expand isothermally and reversibly at 300K from 10 litre to 20 litre, the magnitude of the maximum work obtained is___________J. (nearest integer)

[Given: R=8.3 JK^-1mol^-1 and log2 = 0.3010]

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Payal Gupta

Contributor-Level 10

Number of moles, n = 5 mol

Temperature, T= 300 K

Initial volume, V₁ = 10L

Final volume, V₂ = 20 L

Using;

Work done; w = -2.303 nRT log₁₀ $\frac{V_{2}}{V_{1}}$

$= - 2.303 * 5 * 8 * 300 l o g_{10} \frac{20}{10} J$

= -8630 J

So, magnitude of work done is 8630 J.

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3 months ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

A diatomic gas $(γ = 1.4)$ does 400 J of work when it is expanded isobarically. The heat given to the gas in the process is…………….J.

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Payal Gupta

Contributor-Level 10

$w = P Δ v = n R Δ T = 400 J$ - (1)

$Q = n C_{P} Δ T = \frac{n R y}{y - 1} Δ T$

$Q = \frac{14}{4} * 400$

Q = 1400 J

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3 months ago

Thermodynamics Maths NCERT Exemplar Solutions Class 12th Chapter Twelve

A lead bullet penetrates into a solid object and melts. Assuming that 40% of its kinetic energy is used to heat it, the initial speed of bullet is:

(Given, initial temperature of the bullet = 127°C,

Melting point of the bullet = 327°C,

Latent heat of fusion of lead = 2.5 * 10⁴ J kg^-¹,

Specific heat capacity of lead = 125 J/kg K)

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Payal Gupta

Contributor-Level 10

40% of K.E. = $m c Δ T + m L$

$0.4 * \frac{1}{2} * m v^{2} = m [125 (327 - 127) + 2.5 * 1 0^{4}]$

$\Rightarrow 0.2 v^{2} = [250 * 1 0^{2} + 25 * 1 0^{3}]$

$v^{2} = \frac{2 * 25 * 1 0^{3}}{0.2}$

v = 5 * 100

v = 500 m/sec

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3 months ago

Thermodynamics Physics NCERT Exemplar Solutions Class 12th Chapter Twelve

For a perfect gas, two pressures P1 and P2 are shown in figure. The graph shows:

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Payal Gupta

Contributor-Level 10

At constant Temp

$P α \frac{1}{v}$

v₂ > v₁

⇒ P₁ > P₂

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3 months ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

Assertion (A): A liquid crystallises into a solid and is accompanied by decrease in entropy.

Reason (R): In crystals, molecules organise in an ordered manner.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

option (i)

Explanation: The entropy of a liquid reduces as it crystallises. Because the molecules are more organised in crystalline form.

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3 months ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

Assertion (A): Spontaneous process is an irreversible process and may be reversed by some external agency.

Reason (R): Decrease in enthalpy is a contributory factor for spontaneity.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R is true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

Ans: (ii)

Explanation: The energy factor for a spontaneous process should be favourable (i.e., -ve) and the randomness should be positive.

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3 months ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

Assertion (A): Combustion of all organic compounds is an exothermic reaction.

Reason (R) : The enthalpies of all elements in their standard state are zero.

(i) Both A and R are true and R is the correct explanation of A.

(ii) Both A and R are true but R is not the correct explanation of A.

(iii) A is true but R is false.

(iv) A is false but R is true.

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alok kumar singh

Contributor-Level 10

This is a Assertion and Reason Type Questions as classified in NCERT Exemplar

option (ii)

Explanation: The enthalpy of the reactants is always greater than the enthalpy of the product in a combustion reaction.

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3 months ago

Thermodynamics Chemistry NCERT Exemplar Solutions Class 11th Chapter Six

Match the following:

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alok kumar singh

Contributor-Level 10

This is a Matching Type Questions as classified in NCERT Exemplar

(i)- (b), (d)

(ii)- (b)

(iii)- (c)

(iv)- (a)

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