A gas occupies 11.2L at 0.860kPa. What is the pressure if the volume
becomes 15.0L

A)0.64L
B)1.2kPa
C)0.64kPa
D)1.2L

Option C) 0.64kPa

Explanation:

Given that,

Original volume V1 = 11.2 L

original pressure P1 = 0.860kPa

New volume V2 = 15.0 L

New pressure P2 = ?

Since pressure and volume are involved while temperature is constant, apply Boyle's law

P1V1 = P2V2

0.860 kPa x 11.2 L = P2 x 15.0L

9.632 kPaL = P2 x 15.0L

P2 = 9.632 kPaL / 15.0L

P2 = 0.64 kPa

Thus, the pressure of the gas is 0.64kPa

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.