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2 years ago

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15.0 L of an ideal gas at 298 K and 3.36 atm are heated to 383 K with a new pressure of 5.40 atm. What is the new volume in lite

rs?

1 answer:

klasskru [66]2 years ago

7 0

Key words:-

P=Pressure
V=Volume
T=Temperature

Here

V1=15L
T1=298K
T2=383K
P1=3.36atm
P2=5.40atm
V2=?

Apply combine gas law

$\\ \tt\hookrightarrow \dfrac{P1V1}{T1}=\dfrac{P2V2}{T2}$

$\\ \tt\hookrightarrow \dfrac{3.36(15)}{298}=\dfrac{5.40V2}{383}$

$\\ \tt\hookrightarrow 0.169=0.014V2$

$\\ \tt\hookrightarrow V2=12.1L$

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