This is a simple pairing of two \(\displaystyle Q = mc\Delta T\) equations.

\(\displaystyle mc\Delta T = mc\Delta T\)

We will make the left side for the steel and the right side for water. Thus:

\(\displaystyle (20)(502)(500-90) = (20)(4187)\Delta T\)

Solving for \(\displaystyle \Delta T\) of water yields \(\displaystyle 49.2C\), and certainly the water gained heat not lost it, so the final temperature of the water is \(\displaystyle 70C+49.2C = 119.2C\)

We need to use the equation for thermal expansion in order to solve this problem:

\(\displaystyle \Delta L = \alpha L_o \Delta T\)

We are given:

\(\displaystyle \alpha = 1.2\times 10^{-5}\) for the thermal expansion constant

\(\displaystyle L_o = 7.0 m\) for the initial length of the steel.

We need to calculate \(\displaystyle \Delta T\) which is the the difference of the two temperatures.

\(\displaystyle \Delta T = 38-25 = 13\)

Now we have enough information to solve for the change in the length of steel.

\(\displaystyle \Delta L = (1.2\times10^{-5})(7)(13) = 0.001092m = 1.092mm\)

• We must account for two separate processes since the ice is melting and then warming. The basic formula to calculate the change in entropy at a constant temperature when the ice melts is:

\(\displaystyle \Delta S = \frac{\Delta Q}{T}\)

Where \(\displaystyle \Delta S\) is the change in entropy, \(\displaystyle \Delta Q\) is the change in heat energy, and \(\displaystyle T\) is the temperature at which the change of state for the ice occurs. Now we substitute in an equation for  \(\displaystyle \Delta Q\) for when ice melts: \(\displaystyle \Delta Q = mL_f\). Our entropy equation now becomes:

\(\displaystyle \Delta S = \frac{mL_f}{T}\)

Now we plug in our known values into the equation:

Mass \(\displaystyle m=1kg\)

Latent heat of fusion \(\displaystyle L_f=3.34\times 10^{-5}\frac{J}{kg}\)

Temperature \(\displaystyle T=273.15K\)

And we get the change in entropy for the ice melting at a constant temperature, which is \(\displaystyle 1222.7\frac{J}{K}\)

• Now we need to calculate the change in entropy for the liquid water warming from \(\displaystyle 0C\) to \(\displaystyle 25C\).

The basic formula to calculate the change in entropy at a changing temperature when the water warms is:

\(\displaystyle \Delta S = \int_{T_o}^{T_f} \frac{\Delta Q}{T}\)

Next we substitute in an equation for \(\displaystyle \Delta Q\) when there is a temperature change, but not a change in state. That equation is: \(\displaystyle \Delta Q = mc\Delta T\) so our entropy equation now becomes:

\(\displaystyle \Delta S = \int_{T_o}^{T_f}\frac{mc\Delta T}{T}\)

next we integrate this equation to get:

\(\displaystyle \Delta S = mc \ln(\frac{T_f}{T_o})\)

Next we plug in the values we know and solve for the change in entropy:

Final temperature of water \(\displaystyle T_f = 298.15 K\)

Initial temperature of water \(\displaystyle T_o = 273.15 K\)

Specific heat of water \(\displaystyle c = 4190 \frac{J}{K}\)

Mass of water \(\displaystyle m = 1 kg\)

Now we get the change in entropy for the water warming to be \(\displaystyle 366.9 \frac{J}{K}\)

We add our two entropy values together to find the total change in entropy to be:

\(\displaystyle 1222.7\frac{J}{K} + 366.9\frac{J}{K} \approx 1590\frac{J}{K}\)

In this question, we're told that an ideal gas undergoes a change in its temperature and volume, and we're asked to determine how its pressure changes.

First, we'll need to recall the ideal gas equation.

\(\displaystyle PV=nRT\)

Then, we can rearrange to isolate the term for pressure.

\(\displaystyle P=\frac{nRT}{V}\)

From this expression, we can see that if we double the volume, the pressure will double. Also, cutting the volume in half would double the pressure as well. Therefore, doing both of these would cause the pressure to increase by a factor of \(\displaystyle 4\).

\(\displaystyle P^{'}=\frac{nR(2T)}{0.5\:V}=4\: \frac{nRT}{V}=4P\)

 The first step in this question is to convert the mass of oxygen given into moles.

\(\displaystyle 3.20\:g\:O_2\cdot \frac{1\:mol\:O_{2}}{32\:g\:O_2}=0.10\:mol\:O_{2}\) 

Next, it's important to recognize that we'll need to use the ideal gas equation. Since we're told that the gas is under STP (standard temperature and pressure), we have everything we need to solve for volume.

\(\displaystyle PV=nRT\)

\(\displaystyle V=\frac{nRT}{P}\)

\(\displaystyle V=\frac{(0.10\:mol)(0.08206\:\frac{L\cdot atm}{mol\cdot K})(273\:K)}{1\:atm}=2.24\:L\)

To answer this question, it's important to remember how work is defined when dealing with volume and pressure.

The equation for the work done by a gas on its surrounding is as follows.

\(\displaystyle W=-P\Delta V\)

With this equation in mind, we can see that the only time work is performed is when there is a change in volume. Since we're told in the question stem that the volume remains constant as the pressure is increased, the final volume will be equal to the initial volume. Thus, the \(\displaystyle \Delta V\) term will be equal to \(\displaystyle 0\) and no work will be performed.

In this question, we're told the process that a gas takes through various changes in conditions. From this information, we're asked to find a true statement regarding the work done by the gas in this process.

At its initial point, we can define the gas as having an initial pressure and volume as \(\displaystyle P_{o},\:V_{o}\).

First, we're told that the gas goes through an isothermal expansion to triple its volume. Recall that an isothermal process is one in which the temperature of the gas doesn't change. Moreover, because we're dealing with an ideal gas, we can see from the ideal gas equation that a constant temperature means that a changing volume must by accompanied by a change in pressure. Since the volume is being increased by a factor of three, we know that the pressure must be decreasing by a factor of three. Thus, right after the isothermal expansion has finished, the gas will have a pressure and volume of \(\displaystyle \frac{1}{3}P_{o},\:3V_o\).

\(\displaystyle PV=nRT\)

Next, we're told that the gas undergoes an isobaric contraction back to its original volume. Thus, we know from this that the pressure won't change, but the volume will return to its original value. Consequently, once the isobaric contraction has come to completion, the gas will have a pressure and volume of \(\displaystyle \frac{1}{3}P_{o},\:V_o\).

Finally, we're told that the last process is an isochoric increase of the pressure back to its original amount. Isochoric means that the volume doesn't change in this process. Hence, the gas will now be returned to its original pressure and volume of \(\displaystyle P_{o},\:V_{o}\).

If we were to show this on a pressure vs. volume curve, it would look like the following.

Remember that any pressure-volume work done by a gas is the area under the curve of a pressure vs. volume diagram. For a cyclic process, as the one in this problem, the work done by the gas will be the area within the above image.

For this problem, we don't need to calculate the exact amount of work done. Instead, we can get a rough estimate by first realizing that the structure shown in the above diagram is close to forming a right triangle, but it is slightly smaller. Thus, if we can use the left and bottom sides of the image above to calculate the area that would be enclosed by a right triangle, we can infer that the work done in this problem will be slightly less than the calculated area.

The area for a triangle is the following.

\(\displaystyle A=\frac{1}{2}\: base\cdot height\)

So, for the image shown above, the bottom length is \(\displaystyle 2\:V_{o}\) and the side length is \(\displaystyle \frac{2}{3}\:P_{o}\).  With these values, the area of a right triangle can be calculated.

\(\displaystyle A=(\frac{1}{2})(2\:V_{o})(\frac{2}{3}\: P_{o})=\frac{2}{3}V_{o}P_{o}\)

Finally, since this represents the area for a right triangle, we know that the actual work done is less than this. Hence, the correct answer is:

\(\displaystyle 0< W< \frac{2}{3}V_{o}P_{o}\)

We know that with the ideal gas law that:

\(\displaystyle PV = nRT\)

We also know with isothermal processes, that the work done on a gas is:

\(\displaystyle W = nRT\ln ({\frac{V_f}{V_o}})\)

Since the number of moles and the temperature is not given to us in the question, we can make a substitution with the ideal gas law to get:

\(\displaystyle W = P_oV_o\ln(\frac{V_f}{V_o})\)

Now we can plug in our known values into the equation to solve for the amount of work energy transferred through the isothermal process.

\(\displaystyle W = (1.4\times10^5)(0.50)\ln(\frac{2.2}{0.5}) = 1.04\times10^5\)

Halving the wavelength is equivalent to doubling the frequency since for a light wave the wavelength and frequency are related by \(\displaystyle \lambda\nu = c\), where \(\displaystyle c\) is the speed of light. The energy of a photon is \(\displaystyle E_\text{ph} = h\nu\), where \(\displaystyle h\) is Planck's constant, so the photon energy is doubled.

The photon energy does not change since the color of light (and therefore the frequency of the light) does not change.