Numerical Solutions
Chapter 13: Ideal Gas
From course: Physics Grade XI
Numerical 1
Air at 273 K and 1.01 [latex]\times[/latex]Β 105 Nm-2 pressure contains 2.70 [latex]\times [/latex]Β 1025 molecules per cubic meter. How many molecules per cubic meter will there be at a place, where the temperature is 223K and the pressure is 1.33 [latex]\times[/latex]Β 104 Nm-2?

Solution

Given,

Temperature (T1) = 273 K

Pressure (P1) = 1.01 [latex]\times[/latex]Β 105 Nm-2

No. of molecules per unit volume (n1) = 2.70 [latex]\times[/latex]Β 1025 molecules/m3

No. of molecules per unit volume at T2 (n2) =?

Temperature (T2) = 223 K

Pressure (P2) = 1.33 [latex]\times[/latex]Β 104 Nm-2

We know that,

P = [latex]\frac{1}{3}\rho c^2[/latex]

Or, P = [latex]\frac{1}{3}\times \frac{M}{V}\times c^2[/latex]

Or, P = [latex]\frac{1}{3}\times \frac{mN}{V}\times c^2[/latex]

Or, P = [latex]\frac{1}{3}\times m \times (\frac{N}{V})\times c^2[/latex]

Or, P = [latex]\frac{1}{3}\times m \times n \times c^2[/latex]

So, 1st case:

P1 = [latex]\frac{1}{3}mn_1 \times c^2[/latex] ……………. (i)

For 2nd case:

P2 = [latex]\frac{1}{3}mn_2 [/latex] ……………… (ii)

Dividing eqn. (i) by eqn. (ii), we get,

[latex]\frac{P_1}{P_2} = \frac{\frac{1}{3}mn_1c_1^2}{\frac{1}{3}mn_2c_2^2}[/latex]

Or, [latex]\frac{P_1}{P_2} = \frac{n_1\times T_1}{n_2\times T_2}[/latex]Β Β Β Β Β  [latex][\because c \propto \sqrt{T}][/latex]Β Β Β Β Β Β Β 

Or, n2 = [latex]\frac{n_1\times T_1}{P_1\times T_2}\times P_2[/latex]

= [latex]\frac{(2.70\times 10^{25})\times 273}{(1.01\times 10^5)\times 223}\times (1.33\times 10^4)[/latex]

= 4.353 [latex]\times[/latex]Β 1024 molecules/m3.

Numerical 2
The correct inflation of a tyre at 20oC is 2 kg/cm2. After driving several hours, the driver checks the tyres. If the tyre’s temperature is 50oC, what should be the pressure reading?

Solution

Given,

Temperature (T1) = 20 + 273 = 293 K

Pressure at T1 (P1) = 2kg/cm2

= [latex]2\times 10 \times 10^4[/latex]Β N/m2

= 2 x 105 N/m2

Now, Pressure at T2 (P2) =?

Temperature (T2) = 50 + 273 = 323 K

We know, at constant volume,

[latex]\frac{P_1}{T_1} = \frac{P_2}{T_2}[/latex]

Or, P2 = [latex]\frac{P_1}{T_1}\times T_2[/latex]

= [latex]\frac{2\times 10^5}{293}\times 323[/latex]

= 2.205 x 105 Nm-2.

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