All categories

2 years ago

How is volume and temperature related?

1 answer:

5 0

Charles' Law: The Temperature-Volume Law. This law states that the volume of a given amount of gas held at constant pressure is directly proportional to the Kelvin temperature. As the volume goes up, the temperature also goes up, and vice-versa.

You might be interested in

Why you so sussy bakugou?

ruslelena [56]

Answer:

IM CRYIFNNNAEWJN

Explanation:

6 0

2 years ago

Read 2 more answers

The drawing shows an adiabatically isolated cylinder that is divided initially into two identical parts by an adiabatic partitio

Sveta_85 [38]

Answer:

temperature on left side is 1.48 times the temperature on right

Explanation:

GIVEN DATA:

$\gamma = 5/3$

T1 = 525 K

T2 = 275 K

We know that

$P_1 = \frac{nRT_1}{v}$

$P_2 = \frac{nrT_2}{v}$

n and v remain same at both side. so we have

$\frac{P_1}{P_2} = \frac{T_1}{T_2} = \frac{525}{275} = \frac{21}{11}$

$P_1 = \frac{21}{11} P_2$ ..............1

let final pressure is P and temp $T_1 {f} and T_2 {f}$

$P_1^{1-\gamma} T_1^{\gamma} = P^{1 - \gamma}T_1 {f}^{\gamma}$

$P_1^{-2/3} T_1^{5/3} = P^{-2/3} T_1 {f}^{5/3}$ ..................2

similarly

$P_2^{-2/3} T_2^{5/3} = P^{-2/3} T_2 {f}^{5/3}$ .............3

divide 2 equation by 3rd equation

$\frac{21}{11}^{-2/3} \frac{21}{11}^{5/3} = [\frac{T_1 {f}}{T_2 {f}}]^{5/3}$

$T_1 {f} = 1.48 T_2 {f}$

thus, temperature on left side is 1.48 times the temperature on right

6 0

2 years ago

Which of these does NOT result in the acceleration of an object?

Zanzabum

I guess the correct answer is the first one.

6 0

2 years ago

Read 2 more answers

How much heat is needed to raise the temperature of a 100.0g of water by 85.0 c?

KatRina [158]

Answer:

35.7kJ

Explanation:

we can calculate the amount of heat energy required , using this formula

Q = mcθ

where.

Q = heat energy (Joules, J)

m = mass of a substance (kg)

c = specific heat capacity (units $Jkg^{-1} C^{-1}$ )

θ = change in temperature (Celcius,C or Kelvin K)

Assume Specific heat capacity (c) of water = $4200Jkg^{-1} C^{-1}$

mass =0.1 kg

$Q=0.1 kg*4200Jkg^{-1} C^{-1}*85C\\=35700J\\=35.7kJ$

3 0

2 years ago

Please need help with this

PtichkaEL [24]

Through is the answer to your question

3 0

3 years ago

Read 2 more answers