Physics Thermodynamics

Get insights from 156 questions on Physics Thermodynamics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Thermodynamics

156
Questions
0
Discussions
0
Active Users
1
Followers

Have a question about Physics Thermodynamics? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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

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

$W_{B C} = 0$

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

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

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

Upvote

New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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

Upvote

New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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

Upvote

New answer posted•a year ago

Physics Thermodynamics Class 11th

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…………

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

Efficiency in the Carnot engine is defined as :

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

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

Upvote

New answer posted•a year ago

Physics Thermodynamics Class 11th

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

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

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

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

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

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

Upvote

New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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

$Δ W = n R Δ T - - (i i)$

$\frac{5}{2} = \frac{x}{10} \Rightarrow x = 25.00$

Upvote

New answer posted•a year ago

Physics Thermodynamics Class 11th

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

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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

Upvote

New answer posted•a year ago

Physics Thermodynamics Class 11th

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)

4 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$Δ u = n C_{V} Δ T$

$= n \frac{3 R}{2} Δ T$

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

$= 3486 J$

Upvote

New answer posted•a year ago

Physics Thermodynamics Thermodynamics

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

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

Upvote

New answer posted•a year ago

Physics Thermodynamics Thermodynamics

One mole of an ideal gas at 27°C is taken from A to B as shown in the given PV indicator diagram. The work done by the system will be________× 10^-1 J.

[Given : R = 8.3J/mole K, $l n 2$  = 0.6931]

(Round off to the nearest integer)

15 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Since process is isothermal, so

$W = n R T l n (\frac{V_{f}}{V_{i}}) = 1 \times 8.3 \times 300 \times l n (\frac{4}{2}) = 1 \times 8.3 \times 300 \times l n (2)$

$\Rightarrow W = 1 \times 8.3 \times 300 \times 0.6931 \approx 1725.82 J \approx 17258 \times 1 0^{- 1} J$

Upvote

Physics Thermodynamics

Questions

Discussions

Active Users

Followers