Question

Problem 10.3 A sample of N,
was placed in a flexible 9.0 L container
at 300K at a pressure of 1.5 am The
gas was compressed to a volume of
3.0L and heat was added until the
temperature reached 600K. What is the
new pressure inside the container?
Given Data​

Answer 1

Answer:

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}

T

1

P

1

V

1

=

T

2

P

2

V

2

where,

P_1P

1

= initial pressure of gas = 1.5 atm

P_2P

2

= final pressure of gas = ?

V_1V

1

= initial volume of gas = 9.0 L

V_2V

2

= final volume of gas = 3.0 L

T_1T

1

= initial temperature of gas = 300 K

T_2T

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}

300K

1.5atm×9.0L

=

600K

P

2

×3.0L

P_2=9atmP

2

=9atm