Question

$<i>$One litre of ideal gas is compressed isothermally at 0.72 m of Hg-column so that it's volume becomes 0.9 litre. Find its stress, if the density of mercury is 13.6 × 10³ kg/m³.

[ Answer = 1.07 × 10⁴ N/m² ]
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$\mathsf{Class \ 11th}$
$\mathsf{Chapter \ : \ Mechanical \ Properties \ of \ Solids}$

Answer 1

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$<i>$


Given $\rightarrow$

$\mathsf{V_1}$ = 1 litre

$\mathsf{P_1}$ = 0.72 m of Hg-column

$\mathsf{V_2}$ = 0.9 litre

$\mathsf{P_2}$ = ?



As,

$\mathsf{P_1V_1 \ = \ P_2V_2}$

•°•

$\mathsf{P_2 \ = \ ( \ P_1V_1 \ ) \ / \ V_2}$

Putting values

$\mathsf{P_2}$ = ( 0.72 × 1 ) / 0.9

$\mathsf{P_2}$ = 0.8 m of Hg-column



Now,

Stress = Increase in Pressure

= $\mathsf{P_2 \ - \ P_1}$

= 0.8 - 0.72

= 0.08 m of Hg-column

= 0.08 × ( 13.6 × 10³ ) × 9.8 N / m²

= 10.66 × 10³ N / m²

= 1.07 × 10⁴ N / m²



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Answer 2

hope it help u ..........