Ask question

Ask question

All categories

3 months ago

12

un tanque de gasolina de 40 litros fue llenado por la noche, cuando la temperatura era de 68 grados farenheit. Al dia siguiente

el sol habia elevado la temperatura habia elevado los 131 fareinheit ¿cuanta gasolina derramo el tanque?

1 answer:

inna [3.1K]3 months ago

3 0

Answer:

The gasoline spillage amounts to 1.33 ltr

Explanation:

It is understood that increasing the temperature of a liquid leads to an increase in its volume, defined by

$\Delta V = V_o\gamma \Delta T$

Here, we know that

$V_o = 40 Ltr$

the coefficient of volume expansion for gasoline is specified as

$\gamma = 950 × 10^{–6}$

The temperature variation is described as

$\Delta T = (131 - 68) \times \frac{5}{9}$

$\Delta T = 35 ^oC$

Now we have

$\Delta V = 40(950 \times 10^{-6})(35)$

$\Delta V = 1.33 Ltr$

You might be interested in

Many kinds of analysis in the physical sciences are helped by identifying quantities that are constant. It is usual to begin by

serg [3582]

In circuit analysis, various essential physical quantities must be accurately measured to define circuit characteristics. These include current, denoted as I, representing the flow of electrons, and voltage, which indicates potential difference within the circuit. Most crucially, the resistance, R, outlines the circuit parameters. All these measurements can be determined using a digital multimeter or DMM.

8 0

2 months ago

what is a possible unit for the product VI, where V is the potential difference across a resistor and I is the current through t

Sav [3153]

Recall this formula for a device operating in a direct current circuit:
P = IV
In this equation, P stands for the power emitted by the device, I signifies the current passing through the device, and V represents the voltage drop across it.

Using ampere for current and volt for voltage means that multiplying current by voltage gives you power measured in watts.

5 0

3 months ago

An ideal gas is allowed to expand isothermally from 2.00 l at 5.00 atm in two steps:

Sav [3153]

Heat supplied to the gas = Q = 743 Joules

Work applied to the gas = W = -743 Joules

$\texttt{ }$

Additional explanation

The Ideal Gas Law that should be remembered is:

$\large {\boxed {PV = nRT} }$

P = Pressure (Pa)

V = Volume (m³)

n = number of moles (moles)

R = Gas Constant (8.314 J/mol K)

T = Absolute Temperature (K)

Now, let’s proceed with the problem!

$\texttt{ }$

Given:

Initial volume of the gas = V₁ = 2.00 L

Initial pressure of the gas = P₁ = 5.00 atm

Unknown:

Work done on the gas = W =?

Heat supplied to the gas = Q =?

Solution:

Step A:

An ideal gas expands isothermally:

$P_1V_1 = P_2V_2$

$5.00 \times 2.00 = 3.00 \times V_2$

$V_2 = 10 \div 3$

$V_2 = 3\frac{1}{3} \texttt{ L}$

$\texttt{ }$

Next, we will determine the work performed on the gas:

$W_A = -P_2(V_2 - V_1)$

$W_A = -3.00(3\frac{1}{3} - 2.00)$

$W_A = \boxed{-4 \texttt{ L.atm}}$

$\texttt{ }$

Step B:

By utilizing the methodology mentioned earlier:

$P_2V_2 = P_3V_3$

$3.00 \times 3\frac{1}{3} = 2.00 \times V_3$

$V_3 = 10 \div 2$

$V_3 = 5 \texttt{ L}$

$\texttt{ }$

Next, we will ascertain the work completed on the gas:

$W_B = -P_3(V_3 - V_2)$

$W_B = -2.00(5 - 3\frac{1}{3})$

$W_B = \boxed{-3\frac{1}{3} \texttt{ L.atm}}$

$\texttt{ }$

Ultimately, we can calculate the total work done and heat supplied as follows:

$W = W_A + W_B$

$W = -4 + (-3\frac{1}{3})$

$W = -7\frac{1}{3} \texttt{ L.atm}$

$W = -7\frac{1}{3} \times 101.33 \texttt{ J}$

$\boxed{W \approx -743 \textt{ J}}$

$\texttt{ }$

$\Delta U = Q + W$

$0 = Q + (-743)$

$\boxed{Q = 743 \texttt{ J}}$

$\texttt{ }$

Learn more

Minimum Coefficient of Static Friction:
The Pressure In A Sealed Plastic Container:
Effect of Earth’s Gravity on Objects:

$\texttt{ }$

Answer details

Grade: High School

Subject: Physics

Chapter: Pressure

5 0

3 months ago