Question

If the Pressure and Temperature in the II condition are 2 atm and 600K, what is the volume (V) of the gas? *

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Answer 1

When the balloon is punctured,

The gas in it expands in the large container

Hence,

It’s volume increases and according to Boyle’s law PV=constant. So,

Volume increase, Pressure Decreases.

T

PV

Is constant according to combined gas law, temperature remains constant.

w=−q

ΔV=0

ΔU=q+w=0

Therefore temperature remains constant.

This is the required answer.

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