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An ideal gas in a cubical box having sides of length L exerts a pressure p on the walls of the box. If all of this gas is put into a box having sides of length 4L without changing its temperature, the pressure it exerts on the walls of the larger box will be

A. 1/16 p
B. p
C. 1/64 p
D. 1/4 p

Respuesta :

Answer:

C). [tex]P_2 = \frac{P}{64}[/tex]

Explanation:

As we know if temperature of the gas remains constant then

[tex]P_1V_1 = P_2V_2[/tex]

now we will have

P = initial pressure due to gas

initial volume of the gas is

[tex]V_1 = L^3[/tex]

now we know that side of the box is increased by 4 times

so new volume is

[tex]V_2 = (4L)^3 = 64 L^3[/tex]

now we have

[tex]P(L^3) = P_2(64L^3)[/tex]

now we have

[tex]P_2 = \frac{P}{64}[/tex]

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