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Thermodynamics Part 1

Section: Physics

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Question : 33 of 100

Marks: +1, -0

The temperature of an open room of volume

30\,\mathrm{m}^3

increase from 17°C to 27°C due to sunshine.The atmospheric pressure in the room remains

1\times 10^5

n_i

n_f

are the number of molecules in the room before and after heating then

n_f-n_i

$2.5\times 10^{25}$
$-2.5\times 10^{25}$
$-1.61\times 10^{23}$
$1.38\times 10^{23}$

Solution:

Using the ideal gas equation

PV = NRT

(N is number of moles)

P_0V_0=N_iR\times 290

.....(1)

[T_i=273+17=290\,\text{K}]

After heating

P_0V_0=N_fR\times 300

.....(2)

[T_f=273+27=300\,\text{K}]

from equation (1) and (2)

N_f-N_i=\frac{P_0V_0}{R\times 300}-\frac{P_0V_0}{R\times 290}

difference in number of moles

-\frac{P_0V_0}{R}\left[\frac{10}{290\times 300}\right]

Hence

n_f-n_i

is

=\frac{P_0V_0}{R}\times\left[\frac{10}{290\times 300}\right]\times 6.023\times 10^{23}

putting

P_0=10^5P_A

and

V_0-30\,\mathrm{m}^3

Number of molecules

n_f-n_i=-2.5\times 10^{25}

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