Aflexible container has 5.00 l of nitrogen gas at 298 k. if the temperature is increased to 333k, what will the new volume of that sample of nitrogen?
0.179 l
0.223 l
4.47 l
5.59 l

Givens

Formula

V1/T1 = V2/T2

Note: this will only work if the pressure is constant.

Solution

5.00L / 298 K = x / 333 Multiply both sides by 333 K

5.00 * 333 / 298 = x 333/ 333

5.59 L = V2

5.59 L

Got It right on quiz

Explanation:

V1 = 5.00 L

V2 = ?

T1 = 298k

T2 = 333k

Formula

V1/T1 = V2/T2

Note: this will only work if the pressure is constant.

Solution

5.00L / 298 K = x / 333 Multiply both sides by 333 K

5.00 * 333 / 298 = x 333/ 333

5.59 L = V2

- lead + phosphoric acid = hydrogen + lead(ii) phosphate