Question

A vessel is filled with gas at atmospheric temperature T and pressure P. The mass of the gas is increased by 40% by letting more gas into the vessel at the same temperature. The final pressure of the gas is:

Answer 1

Answer:

We know that

T

1

P

1

= constant for constant volume.

Let the initial pressure be P

1

and initial temperature be T

1

.

Thus final pressure P

2

=1.004P

1

and final temperature T

2

=T

1

+1.

Using

T

1

P

1

=

T

2

P

2

So,

T

1

P

1

=

T

1

+1

1.004P

1

Or T

1

+1=1.004T

1

Or 0.004T

1

=1

⇒T

1

=250K

Answer 2

Answer:

The final pressure of the gas becomes 1.4 times the initial pressure of the gas.

Explanation:

Here we have been given that a vessel is filled with the gas at the atmospheric temperature of T and the pressure P. The mass of the particular gas has increased by 40 % as more gas has been let inside the vessel at the same given temperature.

Now as we know from the ideal gas equation,

$PV = nRT$

Where P is the pressure of the gas, V is the volume, R is the gas constant, T is temperature and n is the number of moles of gas.

As we have been given that only the mass of the gas has been changed remaining all other quantities constant therefore we get,

P ∝ n

as T, V, R is constant here.

Now we know that,

n = Given mass(m) / molar mass of the particular gas(M)

Here as we can see the molar mass of the gas is also constant hence we get,

∴ n ∝ m

∴ P ∝ m

Let $P_{1}$$P_{2}= 1.4 P_{1}$ be the initial pressure of the gas and $P_{2}$ be the final pressure of the gas. Let $m_{1}$ be the initial mass of the gas and $m_{2}$ be the final mass of the gas so we have,

$\frac{P_{1} }{P_{2} } = \frac{m_{1} }{m_{2} }$

Now we have been mentioned that the mass of the gas has been increased by 40% therefore,

∴ $m_{2} = m_{1}$ + 40% of $m_{1}$

⇒ $m_{2} = m_{1} + \frac{40}{100} m_{1}$

⇒ $m_{2} = m_{1} + 0.4m_{1}$

⇒ $m_{2} = 1.4 m_{1}$

⇒ $\frac{m_{2} }{m_{1}} = 1.4$

⇒ $\frac{m_{1} }{m_{2}} = \frac{1}{1.4}$ (equation 1)

∴ $\frac{P_{1} }{P_{2} } = \frac{m_{1} }{m_{2} }$

⇒ $\frac{P_{1} }{P_{2} } = \frac{1}{1.4}$

⇒ $P_{2}= 1.4P_{1}$

Hence the final pressure becomes 1.4 times greater than the initial pressure.