Question

When \[1\text{ }mole\] of gas is heated at constant volume, temperature is raised from\[298\text{ }K\text{ }to\text{ }308\text{ }K\] . Heat supplied to gas is \[500\text{ }J\] . Then which statement is correct?
A ) \[q=W=500J,\text{ }\Delta U=0\]
B ) \[q=\Delta U=500J,\text{ }W=0\]
C ) \[q=\Delta U=-500J,\text{ }W=0\]
D ) \[\Delta U=0,\text{ }q=W=-500J\text{ }\]

Answer 1

Hint: According to the first law of thermodynamics, energy can neither be created nor be destroyed. However energy can be changed from one form into another form. Mathematically, the first law of thermodynamics is given by the following expression.
\[\Delta U = q + w\]

Complete answer:
At constant volume, the initial volume is equal to the final volume. There is no change in volume. \[\Delta V=0\]
The work done is related to the pressure and the volume change through the following relationship:
\[W=P\Delta V\]
When the volume change is zero,
W=P$\Delta V$
W=P(0)
W=0
Hence, the work done at constant volume is zero.
According to the first law of thermodynamics, energy can neither be created nor be destroyed. However energy can be changed from one form into another form. Mathematically, the first law of thermodynamics is given by the following expression.
\[\Delta U=q+w\]
But work done at constant volume is zero.
Hence,
$\Delta U$ = q+w
$\Delta U$=q+0
$\Delta U$=q
Heat supplied to the gas is \[500\text{ }J\] . Heat gained by the system is positive and heat given out by the system is negative.
\[\Delta U=q=+500\text{ J}\]

Hence, the option B ) is the correct answer.

Note: Heated gained by system is positive. The Heat lost by the system is negative. Work done by the surrounding and on the system is positive. Work done by the system is negative.