Tags Physics Thermodynamics

Physics Thermodynamics

Get insights from 155 questions on Physics Thermodynamics, answered by students, alumni, and experts. You may also ask and answer any question you like about Physics Thermodynamics

Follow Ask Question

155

Questions

Discussions

Active Users

Followers

New answer posted

2 months 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)

0 Follower 3 Views

Vishal Baghel

Contributor-Level 10

Since process is isothermal, so

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

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

0 0

New answer posted

2 months ago

Physics Thermodynamics Class 11th

The entropy of any system is given by S = $α^{2} β l n [\frac{μ k R}{J β^{2}} + 3]$ Where a and b are the constants. $μ$ , J, k and R no. of moles, mechanical equivalent of heat, Bolzmann constant and gas constant respectively. $[T a k e S = \frac{d Q}{T}]$

Choose the incorrect option from the following:

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

N_a -> Avogadro's Number

$J β^{2} \equiv [μ k R] \Rightarrow β^{2} \equiv [k R] \Rightarrow β \equiv [k] \equiv [M L^{2} T^{- 2} K^{- 1}]$

, and

-> $S \equiv α^{2} β \Rightarrow [M L^{2} T^{- 2} K^{- 1}] \equiv α^{2} [M L^{2} T^{- 2} K^{- 1}] \Rightarrow α \equiv [M^{0} L^{2} T^{0} K^{0}]$ Dimensionless

0 0

New answer posted

2 months ago

Physics Thermodynamics Class 11th

In the reported figure, hear energy absorbed by a system in going through a cyclic process is_______ pJ.

0 Follower 1 View

alok kumar singh

Contributor-Level 10

Heat absorbed in cyclic process = Work done = 100? Joule

0 0

New answer posted

2 months ago

Physics Thermodynamics Class 11th

The amount of heat needed to raise the temperature of 4 moles of a rigid diatomic gas from 0°C to 50°C when no work is done is___________. (R is the universal gas constant)

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Since process is isochoric, so

$Q = n C_{v} Δ T = n (\frac{f}{2} R) Δ T = 4 (\frac{5}{2} * R) (50) = 500 R$

0 0

New answer posted

2 months ago

Physics 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)

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

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

$Δ W = n R Δ T - - (i i)$

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

0 0

New answer posted

2 months ago

Physics Thermodynamics Thermodynamics

A sample of gas with $γ = 1.5$ is taken through an adiabatic process in which the volume is compressed from 1200 cm³ to 300 cm³. If the initial pressure is 200 kPa. The absolute value of the work done by the gas in the process =_________J.

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Adiabatic equation : $P_{1} V^{γ} = P_{2} V^{γ} \Rightarrow (200) {(1200)}^{γ} = P {(300)}^{γ}$

$\Rightarrow P = (200) {(4)}^{\frac{3}{2}} = 1600 K P a$

$W = \frac{P_{1} V_{1} - P_{2} V_{2}}{γ - 1} = \frac{240 - 480}{1.5 - 1} = - 480 J$

0 0

New answer posted

2 months ago

Physics Thermodynamics Thermodynamics

A mixture of hydrogen and oxygen has volume 500 cm³, temperature 300K, pressure 400 kPa and mass 0.76g. The ratio of masses of oxygen to hydrogen will be:-

0 Follower 2 Views

Vishal Baghel

Contributor-Level 10

Using Ideal gas Equation :

PV = nRT

$\Rightarrow n = \frac{P V}{R T} = \frac{400 * 1 0^{3} * 500 * 1 0^{- 6}}{8.3 * 100} = 0.008$

=> $m_{1} = 0.12, m_{2} = 0.64$

=> $\frac{m_{2}}{m_{1}} = \frac{16}{3}$

0 0

New answer posted

2 months ago

Physics Thermodynamics Class 11th

For an ideal gas the instantaneous change in pressure 'p' with volume 'v' is given by the equation $\frac{d p}{d v} = - a p .$ If p = p₀ at v = 0 is the given boundary condition, then the maximum temperature one mole of gas can attain is:

(Here R is the gas constant)

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

$\frac{d P}{d V} = - a P, a t V = V_{0}, P = P_{0}$

T_max =? for n = 1

$\frac{d P}{d V} = - a P$ (Given)

On integrating, we get

$\int_{P_{0}}^{P} \frac{d P}{P} = - a \int_{0}^{v} d V$

$T_{m a x} = \frac{P_{0}}{a e R} ? n = 1$

0 0

New answer posted

2 months ago

Physics Thermodynamics Class 11th

A heat engine operates between a cold reservoir at temperature T₂ = 400K and a hot reservoir at temperature T₁. It takes 300J of heat from the hot reservoir and delivers 240J of heat to the cold reservoir in a cycle. The minimum temperature of the hot reservoir has to be________K.

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

Kindly go through the solution

$\frac{T_{1}}{400} = \frac{300}{240} \Rightarrow T_{1} = 500 K$

0 0

New answer posted

2 months ago

Physics Thermodynamics Class 11th

The ratio of species heats $(\frac{C_{P}}{C_{V}})$ in terms of degree of freedom (f) is given by:

0 Follower 2 Views

Raj Pandey

Contributor-Level 9

$\frac{C_{P}}{C_{v}} = y = 1 + 2 / f$

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Physics Thermodynamics

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience