Chapter 6, Problem 6.104P

(a)

Interpretation Introduction

Interpretation:

The work done by the expanding gas is to be calculated.

Concept introduction:

The ideal gas equation can be expressed as follows,

$PV=nRT$        (1)

Here,

$P$ is the pressure.

$V$ is the volume.

$T$ is the temperature.

$n$ is the mole of the gas.

$R$ is the gas constant.

The conversion factor to convert $°\text{C}$ to Kelvin is as follows:

$T\left(\text{K}\right)=T\left(°\text{C}\right)+273$        (2)

Work $\left(w\right)$ is the energy needed to move an object in the opposite direction of the force applied. The work done by the system is negative and the work done on the system is positive. The formula to calculate the work done against an external pressure is as follows:

$w=-P\Delta V$        (3)

Here,

$w$ is the work done by the system or on the system.

$P$ is the external pressure.

$\Delta V$ is the change in the volume.

The formula to calculate $\Delta V$ of the system is as follows:

$\Delta V=V_{\text{Final}}-V_{\text{Initial}}$        (4)

The conversion factor to convert $\text{atm}\cdot \text{L}$ into $\text{J}$ is as follows:

$\text{1}\text{atm}\cdot \text{L}=101.3\text{J}$

(b)

Interpretation Introduction

Interpretation:

The temperature change if the gas is heated using the same amount of energy in a container of the fixed volume is to be calculated.

Concept introduction:

Specific heat capacity $\left(c\right)$ of a substance is the quantity to measure the heat needed to raise the temperature of $1\text{g}$ of a substance by $1\text{K}$. The formula to calculate heat required is as follows:

$q=\left(\text{mass}\right)\left(c\right)\left(\Delta T\right)$        (7)

Here,

$\Delta T$ is the temperature difference.

$q$ is the heat released or absorbed.

$c$ is the specific heat capacity of the substance.