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2 years ago

V1T2 = V2T1 is an expression of who’s law

Chemistry

2 answers:

mihalych1998 [28]2 years ago

8 0

Answer:

It is an expression of Charle's law.

Explanation:

Charles' law:

It is one of the gas law.

It states that at constant pressure and for a given moles of a gas, the volume of the gas is directly proportional to the temperature (K).

If we increase the temperature of an ideal gas then at constant pressure and mole, the volume will increase linearly.

V α T

for two volumes at two different temperature

V₁ α T₁

And

V₂ α T₂

diving the two

V₁/V₂ = T₁/T₂

V₁T₂=V₂T1

Andru [333]2 years ago

6 0

<u>Answer:</u>

It is the expression of Charles' Law.

<u>Explanation:</u>

The given expression V1T2 = V2T1 is the formula for the Charles' Law.

A special case of an ideal gas is named as the Charles' Law. This law applies to ideal gases only which are at constant pressure.

According to this law, the volume of a fixed mass of a gas is directly proportional to its temperature and is given by:

V1T2 = V2T1

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2.104 L fuel

Explanation:

Given that:

Volume of water = 35 L = 35 × 10³ mL

initial temperature of water = 25.0 ° C

The amount of heat needed to boil water at this temperature can be calculated by using the formula:

$q_{boiling} = mc \Delta T$

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specific heat of water c= 4.18 J/g° C

$q_{boiling} = 35 \times 10^{3} \times \dfrac{1.00 \ g}{1 \ mL} \times 4.18 \ J/g^0 C \times (100 - 25)^0 C$

$q_{boiling} = 10.9725 \times 10^6 \ J$

Also; Assume that the fuel has an average formula of C7 H16 and 15% of the heat generated from combustion goes to heat the water;

thus the heat of combustion can be determined via the expression

$q_{combustion} =- \dfrac{q_{boiling}}{0.15}$

$q_{combustion} =- \dfrac{10.9725 \times 10^6 J}{0.15}$

$q_{combustion} = -7.315 \times 10^{7} \ J$

$q_{combustion} = -7.315 \times 10^{4} \ kJ$

For heptane; the equation for its combustion reaction can be written as:

$C_7H_{16} + 11O_{2(g)} -----> 7CO_{2(g)}+ 8H_2O_{(g)}$

The standard enthalpies of the products and the reactants are:

$\Delta H _f \ CO_{2(g)} = -393.5 kJ/mol$

$\Delta H _f \ H_2O_{(g)} = -242 kJ/mol$

$\Delta H _f \ C_7H_{16 }_{(g)} = -224.4 kJ/mol$

$\Delta H _f \ O_{2{(g)}} = 0 kJ/mol$

Therefore; the standard enthalpy for this combustion reaction is:

$\Delta H ^0= \sum n_p\Delta H^0_{f(products)}- \sum n_r\Delta H^0_{f(reactants)}$

$\Delta H^0 =( 7 \ mol ( -393.5 \ kJ/mol) + 8 \ mol (-242 \ kJ/mol) -1 \ mol( -224.4 \ kJ/mol) - 11 \ mol (0 \ kJ/mol))$

$\Delta H^0 = (-2754.5 \ \ kJ - 1936 \ \ kJ+224.4 \ \ kJ+0 \ \ kJ)$

$\Delta H^0 = -4466.1 \ kJ$

This simply implies that the amount of heat released from 1 mol of C7H16 = 4466.1 kJ

However the number of moles of fuel required to burn $7.315 \times 10^{4} \ kJ$ heat released is:

$n_{fuel} = \dfrac{q}{\Delta \ H^0}$

$n_{fuel} = \dfrac{-7.315 \times 10^{4} \ kJ}{-4466.1 \ kJ}$

$n_{fuel} = 16.38 \ mol \ of \ C_7 H_{16$

Since number of moles = mass/molar mass

The mass of the fuel is:

$m_{fuel } = 16.38 mol \times 100.198 \ g/mol}$

$m_{fuel } = 1.641 \times 10^{3} \ g$

Given that the density of the fuel is = 0.78 g/mL

and we know that :

density = mass/volume

therefore making volume the subject of the formula in order to determine the volume of the fuel ; we have

volume of the fuel = mass of the fuel / density of the fuel

volume of the fuel = $\dfrac{1.641 \times 10^3 \ g }{0.78 g/mL} \times \dfrac{L}{10^3 \ mL}$

volume of the fuel = 2.104 L fuel

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A sample of a gas at 25°C has a volume of 150 mL when its pressure is 0.947 atm. What will the temperature of the gas be at a pr

sertanlavr [38]

Answer: The temperature of the gas at a pressure of 0.987 atm and volume of 144mL is $25.16^0C$

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The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas = 0.947 atm

$P_2$ = final pressure of gas = 0.987 atm

$V_1$ = initial volume of gas = 150 ml

$V_2$ = final volume of gas = 144 ml

$T_1$ = initial temperature of gas = $25^0C=(25+273.15)K=298.15K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get:

$\frac{0.947\times 150}{298.15}=\frac{0.987\times 144}{T_2}$

$T_2=298.31K=(298.31-273.15)^0C=25.16^0C$

The temperature of the gas at a pressure of 0.987 atm and volume of 144mL is $25.16^0C$

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