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1 Physics -- First Law of Thermodynamics

A cylinder containing 10.0 moles of mono atomic gas expands from A to B along the path shown in figure i) find the temperature of the gas at point A and the temperature at point B II)what is the change in internal energy of gas iii) find the energy transferred to gas by heat in this pro...

A cylinder containing 10.0 moles of mono atomic gas expands from A to B along the path shown in figure i) find the temperature of the gas at point A and the temperature at point B II)what is the change in internal energy of gas iii) find the energy transferred to gas by heat in this process 

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From the ideal gas equation, the relation pressure P, volume V, and temperature of the gas can be expressed as,

 …… (1)

Here, n is number of moles.

From the above formula, the temperature of the gas at point A in the figure P 12.67 can be taken as,

Substitute  for P,  for V, 10 mol for n, and 8.31 for R.

Therefore, the temperature of the gas at point A is .

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A steam engine intakes 200g of steam at 98 degree celsius per minute and cools it down to 10 degree celsius. Calculate the heat rejected by the steam engine per minute. Latent heat of vaporization ...

Heat rejected during condensation of steam per min = mL

= 200*540

= 108000 cal

Heat rejected during colling per min = msdt

= 200*1*(98-10)

= 17600 cal

Total heat rejected by the steam engine per min = 108000+17600 cal

= 125600 cal

Two moles of an ideal mono-atomic gas go through the cycle abc. For the complete cycle, 800J of heat flows out of the gas. Process ab is at constant pressure, and process bc is at constant volume. ...

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Internal energy of system contains: kinetic energy and intermolecular potential energy.

Water falls from 350 high, find the temperature change assuming that the whole of energy due to fallconverted into heat. Swater = 4200JkgC-1 

Solution:

Given,

height of water (h) = 350 m

mass of water (m) = ?

Let acceleration due to gravity be g = 10 m/s^2

Then, the potential energy of water at height h (PE) = mgh

= m (10) (350)

= (3500 m) Joules

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Question says that the entire PE is converted to Q so,

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or, 3500 = 4200 'dt'

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The amount of heat required to change the temperature of 1 mole of gas through 1 K is defined as molar heat capacity of the gas.


From First Law of Thermodynamics,

dQ = dU + dW

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105 calories of heat is required to raise the temperature of 3 moles of ideal gas at constant pressure from 25 C to 30 C. calculate the amount of heat required to raise the temperature of same gas ...

Solution:

Given,

(dQ)P = 105 calories

no of moles (n) = 3

initial temperature (T1) = 25 + 273 = 298 K

final temperature (T2) = 30 + 273 = 303 K

change in temperature (dT) = 303 - 298 = 15 K

We know,

Cp = (dQ)P / n'dT'

= (105 * 4.2) / (3 * 15)

= (9.8)

And,

Cp - Cv = R

or, (9.8) - (8.31) = Cv

So, Cv = 1.5

Now,

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Solution:

n = 5

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V1 = 50 liter

V2 = 5 liter

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Work done = nRT ln(V2/V1)

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