First-Law Tables | CAIE 9702 Paper 4 Notes

Chapters

Paper 4

Physics9702

First-Law Tables

A table question gives some entries and asks for the rest. Complete one row at a time.

Use the condition, then the first law

Write the values given in the row.
Use the process condition to find any zero.
Find the change in internal energy from the states.
Use the first law to find the missing transfer.

Worked example

Compression followed by cooling

An ideal gas has internal energy $U$ at temperature $T$ . It is compressed to temperature $3T$ while work $W$ is done on it. It is then cooled at constant volume to temperature $2T$ (9702/41/M/J/25 Q4(b)).

For a fixed amount of ideal gas, internal energy is proportional to temperature. During compression it rises from $U$ to $3U$ .
$\Delta U_{\mathrm{compression}}=+2U$
Use the first law for the compression row.
$2U=q+W$
$q=2U-W$
Cooling is at constant volume, so no work is done. Internal energy falls from $3U$ to $2U$ .
$W=0$
$\Delta U=-U$
$q=-U$

Compression

W_{\mathrm{on}}=+W

q=2U-W

\Delta U=+2U

Cooling at constant volume

W=0

q=-U

\Delta U=-U

Your turnquick check

In one row of a first-law table, find q.

\Delta U=-80\ \mathrm{J}

W_{\mathrm{on}}=+30\ \mathrm{J}

Show worked answer

$q=\Delta U-W_{\mathrm{on}}$
$q=-80-30=-110\ \mathrm{J}$
The negative sign means 110 J leaves the gas by heating.