Question

A sample of N2 gas was placed in a flexible 9.0 L container at 300 K at a pressure of 1.5 atm. The gas was compressed to a volume of 3.0 L and heat was added until the temperature reached 600 K. What is the new pressure inside the container ?2​

Answer 1

Answer:

New pressure is 9 atm

Explanation:

Here, P1 =1.5 atm; V1 = 9.0 L; V2 = 3.0 L; T1 = 300 K; T2 = 600 K; P2 = ??

According to combined gas equation,

P1V1/(T1) = P2V2/(T2)

Or, P2= P1V1T2/(T1V2)

P2 = (1.5x9.0x600)/(300x3.0)

P2 = 9 atm

Answer 2

The new pressure inside the container is 9 atm

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 1.5 atm

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = 9.0 L

$V_2$ = final volume of gas = 3.0 L

$T_1$ = initial temperature of gas = 300 K

$T_2$ = final temperature of gas = 600 K

Now put all the given values in the above equation, we get:

$\frac{1.5atm\times 9.0L}{300K}=\frac{P_2\times 3.0L}{600K}$

$P_2=9atm$

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