Question

38 ml of most Nitrogen gas were collected at 27 degree Celsius and 74 6.5 mm pressure calculate the volume of the gas at 0 degree Celsius and 760 mm pressure aqueous tension at 27 degree celsius is to 6.5 mm

Answer 1

Solution: According to the question,

P1 = 746.5 mm, V1 = 38 ml, T1 = 27

degree Celsius = 27+273 K = 300 K, T2 = 0 = 0+273 = 273 K,

V2 = 760mm.

As per gas equation, P1*V1*T1 = P2*V2*T2.

Therefore, 746.5*38*300 = 760*V2*273, on calculating, we get V2 = 41 ml.

Answer 2

The volume of the gas is 33.6 bar.

Given:

$\bold{V_1=38\ ml}$

$\bold{P_1=746.5 - 6.5 =740 \ mmHg = 0.98 \ bar}$

$\bold{P_2=760 \ mmHg=1.01 \ bar}$

$\bold{T_1=27^{\circ}\ C = 273+27^{\circ}\ C=300 \ K}$

$\bold{T_2=0^{\circ}\ C = 273+0^{\circ}\ C=273 \ K}$

To find:

$\bold{V_2=?}$

Formula used:

$\bold{\frac{P_{1} V_{1}}{T_{2}}=\frac{P_{2} V_{2}}{T_{2}}}$

Solution:

Now, the volume is given by the formula:

$\bold{V_{2} = \frac{P_{1} V_{1} T_{2}}{P_{2} T_{1}}}$

On substituting the know values, we get,

$\bold{V_2=\frac{0.98\times38\times273}{1.01\times300}}$

$\bold{V_2=\frac{10166.52}{303}}$

The volume at 0 degree Celsius and 760 mmHg is

$\bold{\therefore V_2=33.6 \ bar}$