Twenty (20.0) mL of argon gas (Ar) is stored at STP. If the temperature and pressure are both doubled, what is the new volume?


PLZZ ANSWER ASAP!!

Respuesta :

Answer:

The new volume is the same as the initial volume, i.e. 20 mL.

Explanation:

To find the new volume we can use the Ideal gas law:

[tex] PV = nRT [/tex]

Where:

P: is the pressure

V: is the volume

R: is the gas constant

n: is the number of moles

T: is the temperature

Initially, we have:

[tex] \frac{P_{i}V_{i}}{T_{i}} = nR [/tex]    (1)

In the final state:  

[tex] \frac{P_{f}V_{f}}{T_{f}} = nR [/tex]    (2)

By equating equation (1) and (2) we have:          

[tex] \frac{P_{i}V_{i}}{T_{i}} = \frac{P_{f}V_{f}}{T_{f}} [/tex]  

[tex] \frac{P_{i}V_{i}}{T_{i}} = \frac{2P_{i}V_{f}}{2T_{i}} [/tex]  

[tex]V_{f} = \frac{2T_{i}*P_{i}*V_{i}}{T_{i}*2P_{i}}[/tex]

[tex] V_{f} = V_{i} [/tex]

[tex] V_{f} = 20.0 mL [/tex]

Therefore, the new volume is the same as the initial volume, i.e. 20 mL.

I hope it helps you!

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