Question

5 dm3

of nitrogen are at a temperature of 293 K and a

pressure of 5.05 × 105

Nm–2. What volume will the

gas occupy under N.T.P. conditions?
i am giving u 100 points so pls answer my question and i will mark him as brainliest

Answer 1

Initial volume, V1 = 2.5 dm3
Initial pressure, P1 = 1.01 × 105 Nm-2
Final pressure, P2 = 1.04 × 105 Nm-2
​Final volume = V2
Temperature remain same, using Boyle's law equation:
P1V1 = P2V2
1.01 × 105 Nm-2 × 2.5 dm3​ = 1.04 × 105 Nm-2 × V2
​V2 = 2.43 dm3

Change in volume = V1 - V2 =  0.07 dm3

Answer 2

Heya....

See her for your answer...

=================

Let's see the Solution..

Initial volume ( V1 ) = 2.5dm^3
Initial pressure ( P1 ) = 1.01*105 Nm^-2
Final pressure =
1.04*105Nm^-2

Just know final volume...

Temperature is same so by using Boyle's equation then...

P1V1 = P2V2
1.01*105dm^3 * 105*V2

Hence we get V2 is 2.43 dm...

Now,,, change in volume will be...
V2 - V1 ..

0.72 is the answer...

-- Be Brainly...