Question

The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40°C respectively. If (1/4)th of the gas is released from the vessel and the temperature of the remaining gas is raised to353°C . The final pressure of the gas is

Answer 1

$\sf{\huge{\underline{\pink{Given :-}}}}$

• The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40°C respectively.

• 1/4th of the gas is released from the vessel and the temperature of the remaining gas is raised to 353°C .

$\sf{\huge{\underline{\orange{To\:Find :-}}}}$

• The final pressure of the gas.

$\sf{\huge{\underline{\red{Answer :-}}}}$

Ideal gas are the imaginary gasses .

It follows the equation of pV = nRT, Where , p = pressure, V = vapour, n = number of moles, R = rate of gas constant & T is the temperature of the given gas.

According to the question,

The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40°C respectively.

• p1 = 720 kPa

1 kPa = 1000 Pa

So, 720 kPa = 720 × 1000

➝ 720,000 Pa

• T1 = 40°C = 273 + 40 = 312 K

• T2 = 353°C = 273 +353 = 625 K

1/4th of the gas is released from the vessel.

M2 is released 1/4 . So , M2 remains 3/4 of M1.

Using Formula, PV = $\frac{M}{Mo}$ RT

Where,

V, R & Mo = K

P ∝ m

P ∝ T

P ∝ mT

$\dfrac{p2}{p1} = \dfrac{m2}{m1} \times \dfrac{t2}{t1}$

Let us consider p2 be x

➝ $\dfrac{720}{x} = \dfrac{(3/4)m1}{m1} \times \dfrac{625}{312}$

➝ $\dfrac{720}{x} = \dfrac{3m1}{4m1} \times \dfrac{625}{312}$

➝ $\dfrac{720}{x} = \dfrac{3}{4} \times \dfrac{625}{312}$

➝ x = $720 \times \dfrac{3}{4} \times \dfrac{625}{312}$

➝ x = $\dfrac{1350000}{1248}$

➝ x = 1081.73 pa

The final pressure of the gas is 1081.73 pa .

Answer 2

Given :

• The pressure and temperature of an ideal gas in a closed vessel are 720 kPa and 40°C respectively.

• If (1/4)th of the gas is released from the vessel and the temperature of the remaining gas is raised to353°C .

To find :

• The final pressure of the gas is

Solution :

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ⇝ \sf \: P1 = 720 Kpa$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf⇝1Kpa = 1000pa$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf⇝720 Kpa = 720 × 1000$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf⇝720000pa$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf⇝T1 = 40℃ = 273 + 40 = 312k$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf⇝T2 = 353℃ = 273 + 353 = 625k$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf⇝Using \: formula , pv = \: \frac{M}{M0} \: RT$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sf⇝V, \: R \: and \: M0 = k$

• P ∝ M

• P ∝ T

• P ∝ MT

$\boxed{ \sf \frac{p2}{p1} = \frac{m2}{m1} \times \frac{t2}{t1} }$

• Let us consider P2 be x

$\boxed{ \sf \frac{720}{x} = \frac{ \frac{3}{4} m1}{m1} \times \frac{625}{312} }$

$\boxed{ \sf\frac{720}{x} = \frac{3m1}{4m1} \times \frac{625}{312} }$

$\boxed{ \sf \frac{720}{x} = \frac{3}{4} \times \frac{625}{312} }$

$\boxed{ \sf \: x = 720 \times \frac{3}{4} \times \frac{625}{312} }$

$\boxed{ \sf \: x = \frac{1350000}{1248} }$

$\fbox \pink{x = 1081.73pa}$