Question

An open vessel at a temperature of 20°C is heated at a constant pressure to 340°C.What fraction of air originally contained in the vessel has gone out?

Answer 1

Explanation:

2

air escaped from vessel,

∴

5

3

air remain is vessel.

P and V constant

n

1

T

1

=n

2

T

2

n

1

(300)=(

5

3

n

1

)T

2

⇒T

2

=500K.

Answered By ME

Answer 2

Answer:

The fraction of air contained in the vessel has gone out will be equal to 320/613.

Explanation:

We have given the initial temperature of the vessel, T₁ = 20°

T₁ = 273 + 20 = 293K

The final temperature of the vessel, = T₂ = 340 + 273 = 613K

Consider that 1 mole of gas in vessel before heating, n₁ = 1

We know the ideal gas equation :

PV = nRT

Here, P₁V₁ = n₁RT₁                                                          ............(1)

Similarly, P₂V₂ = n₂RT₂                                                   .........(2)

Because the pressure remains constant, P₁ = P₂ and volume of vessel is constant as well, V₁ = V₂.

Divide the equation (1) by equation (2);

$\frac{P_1V_1}{P_2V_2} =\frac{n_1RT_1}{n_2RT_2}$

$\frac{n_1}{n_2} =\frac{T_2}{T_1}$

$n_2 =\frac{T_1}{T_2}\times n_1$

$n_2=\frac{293}{613}\times 1$

$n_2=\frac{293}{613}$

After heating, fraction of air present in vessel = $n_2=\frac{293}{613}$

After heating, fraction of air getting out of the vessel = 1 - n₂

$=1-\frac{293}{613}$

$=\frac{613-293}{613}$

$=\frac{320}{613}$

Therefore, the fraction of air contained in the vessel has gone out will be equal to 320/613.

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