A sample of carbon monoxide gas whose mass is 6.37 g is heated from 25 oC to 192 oC at a constant pressure of 4.12 bar.

Calculate q (in J) for this process. I got an answer of ~1110J



Calculate w (in J) for this process.

I am unsure how to calculate work considering there is no volume.



Calculate ΔE (in J) for this process.
I know this is DE= q + w , but I need to find volume, can someone help me out?

Well, since you're stuck on finding the volume for calculating work, let me lighten up the situation with a little joke:

Why don't scientists trust atoms?

Because they make up everything!

Okay, now let's get back to the problem. In this case, since the pressure is constant, we can use the formula for work where:

w = -PΔV

But since there's no change in volume, ΔV = 0. So, the work done in this process would be zero:

w = -PΔV = -4.12 bar * 0 = 0 J

Now, let's calculate ΔE using the formula ΔE = q + w. Since we know that w = 0 J, the equation simplifies to:

ΔE = q

Therefore, the change in internal energy (ΔE) for this process would be the same as the heat transfer (q), which you previously calculated to be ~1110 J.

So, ΔE = q = ~1110 J.

I hope that cleared things up for you! Let me know if you have any more questions or if you need another joke to lighten the mood!

To calculate work (w) for this process, we can use the equation:

w = -PΔV
Here, P represents the constant pressure and ΔV represents the change in volume.

Since no volume information is given, we can assume that the gas behaves ideally and obeys the ideal gas law, PV = nRT.

At constant pressure (P) and temperature (T), the equation simplifies to:
V2 = V1 * (T2 / T1)
where V1 and V2 are the initial and final volumes, and T1 and T2 are the initial and final temperatures.

Since the volume is not given, we can assume an initial volume of 1 L and find the final volume using the ideal gas law. We can use the equation V2 = nRT2 / P, where n represents the number of moles and R is the ideal gas constant (0.0821 L * atm / (mol * K)).

First, we need to find the number of moles using the formula n = m / M, where m is the mass of the gas and M is the molar mass of carbon monoxide (28.01 g/mol).

n = 6.37 g / 28.01 g/mol
= 0.227 mol

Now, we can find the final volume (V2) using the equation:
V2 = nRT2 / P
= (0.227 mol) * (0.0821 L * atm / (mol * K)) * (465 K) / (4.12 atm)
= 5.39 L

To calculate work:
w = -PΔV
= -(4.12 atm) * (5.39 L - 1 L)
= -20.96 L * atm

Note: The unit of atm * L is equivalent to Joules, so we can directly write the answer as:
w = -20.96 J

Now, to find the change in internal energy (ΔE):
ΔE = q + w

We already have the value for q, which is approximately 1110 J, as you stated.

ΔE = 1110 J + (-20.96 J)
= 1089.04 J

Therefore, the change in internal energy (ΔE) for this process is approximately 1089.04 J.

To calculate q (heat) for the given process, you can use the equation:

q = m * C * ΔT

where:
- q is the heat (in J)
- m is the mass of the gas (in g)
- C is the specific heat capacity of the gas (in J/(g·°C))
- ΔT is the change in temperature (in °C)

For carbon monoxide gas, the specific heat capacity (C) is approximately 28.0 J/(g·°C).

Using the given values, we can substitute them into the equation:

q = (6.37 g) * (28.0 J/(g·°C)) * (192 °C - 25 °C) = 1115.79 J (approximately)

Therefore, the value for q is approximately 1115.79 J.

Now, let's move on to calculating w (work):

Since the problem states that the process occurs at constant pressure, we can use the equation:

w = -P * ΔV

where:
- w is the work done on or by the gas (in J)
- P is the pressure (in bar)
- ΔV is the change in volume (in L)

However, as you mentioned, the problem does not provide any information about the volume. Without the volume, it is not possible to calculate the work done on or by the gas.

Finally, to calculate ΔE (change in internal energy), you can use the equation:

ΔE = q + w

Since we determined that q is approximately 1115.79 J and we don't have the value for w, we cannot calculate the precise value for ΔE.

It's a constant volume. Work is =pdeltaV and since delta b is zero, then work is zero.

q = mass CO x specific heat CO x delta T.
You didn't list sp. h. and I don't know all of those by memory.