Question

A gas in a cylinder is held at a constant pressure of 2.3X 10^5 Pa and is cooled and compressed from 1.70 m^3 to 1.20 m^3.  The internal energy of the gas decreases by 1.4 X 10^5 J.  a) Find the work done by gas. b) Find the absolute value of Q of the heat flow into or out of the gas, and state the direction of the heat flow.  c) Does it matter whether the gas is ideal? why or why not?

Answer 1

Concepts and reason

The concept used to solve this problem is first law of thermodynamics.

First use the relation between pressure, final volume, and initial volume to calculate the work done by the gas.

Finally use the first law of thermodynamics to calculate heat flow into or out of the gas and the direction of heat flow.

Fundamentals

First law of thermodynamics states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

$\Delta U = Q - W$

Here, $\Delta U$ is the decrease in internal energy, $Q$ is the heat out of the gas, and $W$ is the work done by the gas.

Rearrange the above equation to get heat out of the gas,

$Q = \Delta U + W$

Expression for the work done by the gas is,

$W = P\left( {{V_{\rm{f}}} - {V_{\rm{i}}}} \right)$

Here, P is the pressure, ${V_{\rm{f}}}$ is the final volume, and ${V_{\rm{i}}}$ is the initial volume.

Expression for the ideal gas law is,

$PV = nRT$

Here, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

(a)

Expression for the work done by the gas is,

$W = P\left( {{V_{\rm{f}}} - {V_{\rm{i}}}} \right)$

Substitute $2.3 \times {10^5}\,{\rm{Pa}}$ for $P$ , $1.70\,{{\rm{m}}^3}$ for ${V_{\rm{i}}}$ , and $1.20\,{{\rm{m}}^3}$ for ${V_{\rm{f}}}$ .

$\begin{array}{c}\\W = \left( {2.3 \times {{10}^5}\,{\rm{Pa}}} \right)\left( {1.20\,{{\rm{m}}^3} - 1.70\,{{\rm{m}}^3}} \right)\\\\ = - 1.15 \times {10^5}\,{\rm{J}}\\\end{array}$

(b)

Expression for heat out of the gas is,

$Q = \Delta U + W$

Substitute $- 1.15 \times {10^5}\,{\rm{J}}$ for $W$ and $- 1.4 \times {10^5}\,{\rm{Pa}}$ for $\Delta U$

$\begin{array}{c}\\Q = \left( { - 1.15 \times {{10}^5}\,{\rm{J}}} \right) + \left( { - 1.4 \times {{10}^5}\,{\rm{Pa}}} \right)\\\\ = - 2.55 \times {10^5}\,{\rm{J}}\\\end{array}$

The heat flow has negative sign, so the heat flow is out of the gas.

(c)

Ideal gas equation not required to calculate the work done by the gas and heat flow. Ideal gas is not used either directly or indirectly.

Hence, it does not matter whether the gas is ideal or not.

Ans: Part a

The work done by the gas is $- 1.15 \times {10^5}\,{\rm{J}}$ .

Part b

The absolute value of Q is $- 2.55 \times {10^5}\,{\rm{J}}$ and the direction of heat flow is out of the gas.

Part c

It does not matter whether the gas is ideal or not.