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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

User Redexp
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1 Answer

5 votes

Answer:

C).
P_2 = (P)/(64)

Step-by-step explanation:

As we know if temperature of the gas remains constant then


P_1V_1 = P_2V_2

now we will have

P = initial pressure due to gas

initial volume of the gas is


V_1 = L^3

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

so new volume is


V_2 = (4L)^3 = 64 L^3

now we have


P(L^3) = P_2(64L^3)

now we have


P_2 = (P)/(64)

User Roldan
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