Question

Part II - A. "Say Yes to Challenges, Yes You Can Manage"
DIRECTIONS: Solve the following problems by using G.R.E.S.A (Given, Required,
Equation, Solution, and Answer). Use extra sheet/s to show your solutions.

1. At constant temperature, the volume of a gas was found to be 500cm3 at a pressure of 760mmHg. If the pressure of the gas is doubled what will be the new volume of the gas?
2. Akila wanted to buy a helium tank to fill up all the balloons she needed for a birthday party. If the 25.00L Helium at 14.70psi will be compressed down to a volume of 2.20L, what
would be the final pressure of the helium?
3. A sample of Neon gas occupies a volume of 0.50L at 1.70atm. What will be its volume at 30kPa?
4. At constant pressure, Hydrogen gas contracts from 1.00L to 0.95L. The initial temperature is 293.15K. Find the final temperature of the gas.
5. Carbon dioxide gas expands from 10.00cm3 to 11.50cm3. The final temperature is 450C. Determine the initial temperature if the expansion is elemental.

Answer 1

Answer:

I do not understand can you tell answer

Answer 2

Answer:

1. $V_{1}$= 500 cm³

P1 = 760 mm Hg

new pressure = 2 P1

We need to find $V_{2}$

The relation between pressure and volume is P1$V_{1}$=P2$V_{2}$.  P1 is original pressure, P2 is final pressure. In addition,$V_{1}$ is original volume and $V_{2}$ is final volume.

Firstly we will find P2

P2= 2 x P1 + P1

P2=3P1

Now substituting the value in the formula P1$V_{1}$=P2$V_{2}$

760 x 500= 3x760x $V_{2}$

2,30,000= 2,280 x $V_{2}$

$V_{2}$=100.87cm³

2. V1= 25 L

V2= 2. 20 L

P1= 14. 70 psi

From the formula,

P2= P1V1/V2

Solution:

(14.70 psi) × (25 L) / (2.20 L ) = 167.04 psi

14.70×25÷2.20= 167.04 psi

3. $P_{1}$ = 1.70 ATM

$V_{1}$= 0.50 L

$P_{2}$ = 30 kPa OR 0.3 atm

$V_{2}$ = ?

We again will use the formula,

$P_{1} * V_{1} = P_{2} * V_{2}$

1.70 * 0.50 = 0.3 * $V_{2}$

$V_{2}$ = 2.83 L

4. Given,

$V_{1}$ = 1.00L

$V_{2}$ = 0.95L

$T_{1}$ = 293.15K

$T_{2}$ = ?

We will use the formula =>

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

$\frac{1}{293.15} = \frac{0.95}{T_{2} }$

$T_{2}$ = 0.95 x 293.15

=278.49 K

5.We have,

$V_{1}$=10 cm³

$V_{2}$=11.50 cm³

$T_{1}$= ?

$T_{2}$= 450°C = 318.15 K

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

$\frac{10}{T_{1} } =\frac{11.50}{318.15}$

$T_{1} = \frac{10* 318.15}{11.50}$

$T_{1} =$ 276.65 K

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