Compressing a gas according to an absurd law!

Classical Mechanics Level 3

Consider a situation in which 1 1 mole of a diatomic gas is compressed to ( 1 4 ) t h (\dfrac{1}{4} )^{th} of its original volume according to the law V T 2 = C VT^2 = C , where C is a constant. If the initial pressure was 1 0 5 10^5 pascal calculate the change in pressure in terms of pascal .

Details

  • Be sure to calculate the change, i.e. Δ P = P f − P i \Delta P = P_f-P_i and not the fold-change P f / P i P_f/P_i .


The answer is 700000.

Kunal Joshi by Kunal Joshi , 24, India

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3 solutions

Kunal Gupta
Dec 1, 2014

now using ideal gas eqn: p V = n R T pV=nRT ; here n = 1 n=1 . now process eqn is: V T 2 = k VT^{2} =k i.e. ( P 2 ) ( V 3 ) = k ′ (P^{2})(V^{3})=k' hence P 2 = 8 P 1 P_{2}= 8P_{1} ; hence change in pressure: 7.0 ∗ 1 0 5 P a 7.0 \quad * 10^{5} Pa

2 Helpful 0 Interesting 0 Brilliant 0 Confused
Anubhav Tyagi
Feb 12, 2016

Good for starters

0 Helpful 0 Interesting 0 Brilliant 0 Confused
Vatsalya Tandon
Feb 11, 2016

V T 2 = C VT^2 = C

V ( P V n R ) 2 = C V(\frac{PV}{nR})^2 = C

P 2 V 3 = C . N 2 R 2 P^2V^3 = C.N^2R^2

L e t t h i s b e K Let \quad this \quad be \quad K

P V 3 2 = K PV^\frac{3}{2} = K

1 0 5 X V 1 . 5 = P ( f i n a l ) X V 1.5 4 1.5 10^5 \quad X \quad V^1.5 = P(final) \quad X \quad \frac{V^{1.5}}{4^{1.5}}

P ( f i n a l ) = 8 X 1 0 5 P(final) = 8 \quad X \quad 10^5

Δ P = 7 X 1 0 5 \Delta P = 7 \quad X \quad 10^5

0 Helpful 0 Interesting 0 Brilliant 0 Confused

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