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Boyle's law states that if the temperature of a gas remains​ constant, then PV​=c, where P=pressure​,V=volume ​, and c is a constant. Given a quantity of gas at constant​ temperature, if V is decreasing at a rate of 10in​^3/sec​, at what rate is P increasing when P =60​lb/in^2 and V=20in^3 in​? (do not round your answer.)

a. 30 lb/in^2 per sec
b. 120lb/in^2 per sec
c. 9 lb/in^2 per sec
d. 10/3 lb/in^2 per sec

Respuesta :

adioabiola

The pressure, P if V is decreasing at a rate of 10in^3/sec is 120lb/in^2 per sec.option B

Boyle's law

PV = c

where,

  • P=pressure,
  • V=volume , and
  • c is a constant

when P =60lb/in^2 and V=20in^3

PV = c

60 × 20 = 1200 in

Find P if V is decreasing at a rate of 10in^3/sec

PV = c

P × 10 = 1200

10P = 1200

P = 1200 / 10

P = 120lb/in^2 per sec

Learn more about Boyle's law:

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