Question

A gas occupies 340cc at STP.Find its volume at 20°C and 700 mm of Hg​

Answer 1

Answer:

V

1

=700 ml

P

1

=1 atm=760 mm Hg

T

1

=298 K V

2

=?

P

2

=400 mm Hg

T

1

=288 K

Here number of moles is constant.

∴

T

1

P

1

V

1

=

T

2

P

2

V

2

V

2

=

T

1

×P

2

P

1

×V

1

×T

2

=

298×400

760×700×288

=1285.37 ml

∴ Volume occupied by gas is 1285.37 ml.

Answer 2

Given:

A gas occupies 340cc at STP

To find:

Find its volume at 20°C and 700 mm of Hg

Solution:

According to Ideal Gas Law, we have

$\boxed{\bold{PV = nRT}}$

since the value of n & R of the given gas will remain constant, so we can rewrite the law as

$\implies \boxed{\bold{\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} }}$ ...... (i)

Here we are given,

V₁ = 340 cc = $\frac{340}{1000}\:L$ = 0.34 L

at STP, P₁ = 1 atm = 760 mmHg

at STP, T₁ = 273 K

P₂ = 700 mmHg = $\frac{700}{760}$ = 0.921 atm

T₂ = 20° C = 20 + 273 K = 293 K

Let "V₂" represents the volume of the gas when its pressure is 700 mmHg and the temperature is 20° C.

Now, on substituting the values in the law written in (i), we get

$\frac{1 \:atm\times \:0.34\:L}{273\:K} = \frac{0.921 \:atm\times \:V_2\:L}{293\:K}$

$\implies V_2 = \frac{1 \:\times \:0.34\:\times\:293}{273\:\times \:0.921}$

$\implies V_2 = \frac{99.62}{251.433}$

$\implies V_2 = \frac{99.62}{251.433}$

$\implies V_2 =0.39620\:L$

$\implies \bold{V_2 = 396.2\:cc}$

Thus, its volume at 20°C and 700 mm of Hg​ is → 396.2 cc.

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