Question

The pressure exerted by a certain mass of enclosed at 300K
Is 5x10^4 what will bethe pressure exterted by the gas at 600 K the volume of the gas is kept constant ?

Answer 1

Answer:

According to Charles's law -

⇒

T

1

V

1

=

T

2

V

2

⇒V

2

=(

T

1

V

1

)×T

2

=450×(

300

1×10

−4

)

=1.5×10

−4

dm

3

Hence, the answer is 1.5×10

−4

dm

3

.

Answer 2

Given:

Pressure at $300K=5$ × $10^{4}N/m^{2}$

To find:

Pressure at $600K$

Solution:

We have been given that the pressure exerted is $5$ × $10^{4}N/m^{2}$ at $300K$.

We have,

Ideal Gas Equation as $PV=nRT$

Where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the Rydberg's constant and $T$ is the temperature.

We have been given that the volume of the gas is constant. Hence,

$P$ ∝ $T$

That is, the temperature of the gas increases as the pressure is increased.

Now,

According to the question, we have,

$P_{1} =5$ × $10^{4}N/m^{2}$      $;$ $T_{1} =300K$     $;$ $T_{2} =600K$

Hence, the pressure at $T_{2}$ will be

$P_{2}=\frac{P_{1} *T_{2} }{T_{1} }$

$P_{2}=\frac{5*10^{4}*600 }{300}$

$P_{2}=10^{5}N/m^{2}$

Final answer:

Hence, the pressure exerted by the gas at 600 K will be 10⁵ N/m².