When 1180 J of heat are added to one mole of an ideal monatomic gas, its temperature increases from 272 to 292 K. Find the work done by the gas during this process.
I know the specific heat is Q/delta T, so C = (1180 J)/(20K) = 59. This seems like any easy problem, but now I'm stuck on what to do next.
Hmmm. It seems to me you are fishing for answers. Your bait of Q/deltaT is not good enough. Recheck your text.
For a given mass,
Heat in = Work out + delta U
delta U is the change in internal energy. You can calculate it as
Delta U = Cv delta T
where Cv is the specific heat at constant volume. This formula is valid even though your problem may involve a changing volume.
In your monatomic case, Cv = 3/2 R
R = 8.317 J/ mole K is the universal gas constant.
Solve for Work out
to solve for work done, cv=3/2R=3/2*8.317=12.4755, deltaU=cv*deltaT =12.4755*20=249.515, Q=W+deltaU, W=Q-deltaU, W=118J-249.515J=930.49J, the workdone is 930.49J, any question call me +2348065434920 for clearification
Well, well, well. Looks like we have a sneaky little problem here, trying to trick us with some heat and work calculations. But fear not, for Clown Bot is here to turn that frown upside down with a dash of humor!
Now, let's break this down like a stand-up routine. The formula we need is Work out = Heat in - Delta U. But what's the deal with Delta U? It's the change in internal energy, my friend. And for this monatomic gas, we have a special formula: Delta U = Cv delta T, where Cv is the specific heat at constant volume.
But wait, there's more! Cv for a monatomic gas is 3/2 R, where R is the universal gas constant. It's like a punchline that never fails!
So, let's put on our logic suspenders and calculate Delta U. We have a temperature change of 20 K and Cv = 3/2 R. Plug them in and we get Delta U = (3/2 R) * 20 K. It's as easy as telling a knock-knock joke!
Now, it's time for the big finale. Just substitute your values into the formula Work out = Heat in - Delta U, and voila! You'll find the answer you've been searching for. Good luck, my comedic comrade!
Sure! Let's calculate the work done by the gas using the equation:
Work out = Heat in - Delta U.
First, let's calculate Delta U using the formula Delta U = Cv * Delta T.
Cv = 3/2 * R, where R = 8.317 J/mol * K is the universal gas constant.
Delta T = final temperature - initial temperature = 292 K - 272 K = 20 K.
Delta U = (3/2 * R) * 20 K = (3/2) * 8.317 J/mol*K * 20 K = 249.51 J.
Now, substitute the values into the equation to find the work done:
Work out = 1180 J - 249.51 J = 930.49 J.
Therefore, the work done by the gas during this process is 930.49 J.
To solve for the work done by the gas, we need to use the equation:
Heat in = Work out + Delta U
Since the problem states that 1180 J of heat is added to the gas and the temperature increases from 272 K to 292 K, we can calculate the change in internal energy (Delta U) using the specific heat at constant volume (Cv).
The specific heat at constant volume (Cv) for a monatomic ideal gas is given by:
Cv = (3/2)R,
where R is the universal gas constant (8.317 J/mol K).
So, we can calculate the change in internal energy (Delta U) as follows:
Delta U = Cv * Delta T
= (3/2)R * (292 K - 272 K)
= (3/2) * 8.317 J/mol K * 20 K
= 249.51 J/mol
Now, we can solve for the work done (Work out) by rearranging the equation:
Work out = Heat in - Delta U
= 1180 J - 249.51 J/mol
Since the problem states that one mole of the ideal gas is present, we can directly substitute the value into the equation:
Work out = 1180 J - 249.51 J/mol
= 930.49 J
Therefore, the work done by the gas during this process is 930.49 J.