Hello!

To solve this problem, we will use the Boyle's Law, which describes how pressure changes when volume changes and vice-versa. The equation for this law is the following one, and we'll clear for V2:

$P1*V1=P2*V2 \\ \\ V2= \frac{P1*V1}{P2}= \frac{3,24 L * 1 atm}{1,20 atm}= 2,7 L$

So, the final volume after increasing the pressure would be 2,7 L. That means that volume decreases when the pressure increases

Have a nice day!

8 0

4 years ago

A sample of gas occupies 10.0 L at 240°C under a pressure of

NISA [10]

Answer: 1090°C

Explanation: According to combined gas laws

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 = initial pressure of gas = 80.0 kPa

V1 = initial volume of gas = 10.0 L

T1 = initial temperature of gas = 240 °C = (240 + 273) K = 513 K

P2 = final pressure of gas = 107 kPa

V2 = final volume of gas = 20.0 L

T2 = final temperature of gas

Substituting the values,

(80.0 kPa × 10.0 L) ÷ (513 K) = (107 kPa × 20.0 L) ÷ T2

T2 = 513 K × (107 kPa ÷80.0 kPa) × (20.0 L ÷ 10.0 L)

T2 = 513 K × (1.3375) × (2)

T2 = 1372.275 K

T2 = (1372.275 - 273) °C

T2 = 1099 °C

8 0

3 years ago

Read 2 more answers

If the visible light spectrum is from 400 to 700 nm, would light with an energy of 2.79 x 10^-19 J be visible with the naked eye

sladkih [1.3K]

Answer:

713 nm. It is not visible with the naked eye.

Explanation:

Step 1: Given data

Energy of light (E): 2.79 × 10⁻¹⁹ J

Planck's constant (h): 6.63 × 10⁻³⁴ J.s

Speed of light (c): 3.00 × 10⁸ m/s

Wavelength (λ): ?

Step 2: Calculate the wavelength of the light

We will use the Planck-Einstein equation.

E = h × c / λ

λ = h × c / E

λ = 6.63 × 10⁻³⁴ J.s × 3.00 × 10⁸ m/s / 2.79 × 10⁻¹⁹ J

λ = 7.13 × 10⁻⁷ m

Step 3: Convert "λ" to nm

We will use the relationship 1 m = 10⁹ nm.

7.13 × 10⁻⁷ m × (10⁹ nm/1 m) = 713 nm

This light is not in the 400-700 nm interval so it is not visible with the naked eye.

5 0

3 years ago