Calculate the volume change with gas temperature for this piston in a cylinder

[Istiak]

Homework Statement

The top of an empty cylinder is covered with a piston. The forces applied to the piston are the atmospheric pressure, the ideal gas pressure, and the gravitational force applied to it due to the mass of the piston. At first the gas pressure was 0. If Q amount of heat is given to the gas slowly, how much will the volume of the gas change? The gas is one atom and the heat holding capacity of cylinders and pistons is negligible

Relevant Equations

U=\frac{f}{2} RT PV=nRT Q=MS \delta T

Solution attempt :

Option :

I am sure that my work is wrong. But, I must add solution attempt in PF that's why I just added that. How can I solve the problem?

 

[haruspex]

Are you not given the cross-sectional area of the cylinder?

 

[Istiak]

Are you not given the cross-sectional area of the cylinder?

Ohh! Sorry! No

 

[haruspex]

Ohh! Sorry! No

Then I see no way to relate the pressure of the gas to the weight of the piston,

 

[Chestermiller]

Your translation of the problem statement doesn't make sense to me.

 

[TSny]

In terms of the number of moles, $n$, of a monatomic gas, you have

$PV = nRT$ and $U = \frac 3 2 nRT$.

It looks like you left out $n$ in the equation for $U$.

Can you combine these two equations to express $V$ in terms of $U$ and $P$?

(I assume that the phrase "the gas is one atom" means each molecule of the gas consists of one atom.)

 

[haruspex]

... and I assume the initial gas pressure of zero is gauge pressure, i.e. absolute pressure is atmospheric.

 

[TSny]

... and I assume the initial gas pressure of zero is gauge pressure, i.e. absolute pressure is atmospheric.

I'm going to guess that the OP meant to type "P0" instead of "0" for the initial pressure.

The pressure of the gas needs to be greater than atmospheric pressure in order to support the piston.

 

[haruspex]

I'm going to guess that the OP meant to type "P0" instead of "0" for the initial pressure.

View attachment 285870

The pressure of the gas needs to be greater than atmospheric pressure in order to support the piston.

yes, that makes more sense.

 

[Istiak]

I'm going to guess that the OP meant to type "P0" instead of "0" for the initial pressure.

View attachment 285870

The pressure of the gas needs to be greater than atmospheric pressure in order to support the piston.

Actually, they wanted to write that the pressure was P_0 at first. Then, they wrote if we use Q amount of heat slowly than, how much volume will change? (In your taken picture there wasn't "will change" word I took it from main picture).

 

[Istiak]

(I assume that the phrase "the gas is one atom" means each molecule of the gas consists of one atom.)

The gas is mono-atomic. I just copy-and-pasted from google translate. That's why it is written `one atom`.

 

[Istiak]

PV=nRT and U=32nRT.

Actually, I found in Internet that $U=\frac{3}{2}RT$. I actually forgot the main equation that's why I searched in internet then, found it. Let me see if I get the same ss for you again.

 

[Istiak]

In terms of the number of moles, $n$, of a monatomic gas, you have

$PV = nRT$ and $U = \frac 3 2 nRT$.

It looks like you left out $n$ in the equation for $U$.

Can you combine these two equations to express $V$ in terms of $U$ and $P$?

(I assume that the phrase "the gas is one atom" means each molecule of the gas consists of one atom.)

 

[TSny]

In the table, $U$ represents the internal energy per mole of gas. $C_P$ and $C_V$ are the heat capacities per mole (called the "molar heat capacities").

For a monatomic gas containing $n$ moles, the total internal energy of the gas is $U = \frac 3 2 nRT$.

 

[Istiak]

In the table, $U$ represents the internal energy per mole of gas. $C_P$ and $C_V$ are the heat capacities per mole (called the "molar heat capacities").

For a monatomic gas containing $n$ moles, the total internal energy of the gas is $U = \frac 3 2 nRT$.

$U=\frac{3}{2} PV$
$Q-W=\frac{3}{2}PV$
$V=\frac{2}{3P}(Q-W)$

I can't move futher.

 

[TSny]

$U=\frac{3}{2} PV$

Good

$Q-W=\frac{3}{2}PV$

$Q-W$ represents the change in internal energy $\Delta U$. So, $Q-W = \Delta U$.

From the equation $U = \frac 3 2 PV$, can you get an equation for $\Delta U$ in terms of $\Delta V$?

Can you express $W$ in terms of $\Delta V$?

Hint: Think about whether or not there will be any change in the pressure of the gas as the heat is slowly added.