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a gas occupies 0.6 litres under a pressure of 0.92 bar. under what pressure will the volume of gas be reduced by 20% of its original volume ? ( temp is constant)
is the answer 4.6?
at a constant temperature P is infersely proportionsl to 1/V PV=comstant at same temperature apply P1V1 = P2V2 thus u will get P as 1.15 bar
at a constant temperature P is infersely proportionsl to 1/V
PV=comstant
at same temperature apply P1V1 = P2V2
thus u will get P as 1.15 bar
P1V1=P2V2.92*.6= .12*P2 P2= 4.6 bar :answer
P1V1=P2V2.92*.6= .12*P2
P2= 4.6 bar :answer
Hi Tejas, V1=0.6 lit V2=V1-V1*20/100 = V1(1-20/100)=V1(80/100) P1=0.92 bar Boyle's law: P1*V1=P2*V2 0.92*V1=P2*V1*0.8 P2=0.92/0.8=1.15 bar
Hi Tejas,
V1=0.6 lit
V2=V1-V1*20/100
= V1(1-20/100)=V1(80/100)
P1=0.92 bar
Boyle's law: P1*V1=P2*V2
0.92*V1=P2*V1*0.8
P2=0.92/0.8=1.15 bar
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