Question

4. The pressure of an ideal gas is doubled during a process in which the energy given up as heat by the gas equals the
work done on the gas. As a result, the volume is:
A) doubled
B) halved
C) unchanged
D) need more information to answer
E) nonsense, the process is impossible


Please Explain.

Answer 1

By the first principle of thermodynamic,
\[\Delta U=W+Q\] As Q (is negative because is given up) equals W, then the process is isotherm, so,
\[W=Q\Rightarrow \Delta T=constant\]Then it is possible to apply,
\[P_1V_1=P_2V_2\Rightarrow P_1V_1=2P_1V_2\Rightarrow V_2=V_1/2\]
And the correct answer is b.

Answer 2

Thanks ,
When do we use Q=U+W and when do we use U=W+Q
What difference between them?

What exactly define Q , U , W?

Answer 3

@John_ES

Answer 4

It all dependes on the sign of the work and the heat transfered. The usual is to put always,
\[\Delta U=Q+W\]
And then, put the signs. For example, for a process that gives heat, Q-> -Q, and do work against the exterior W>0, so,
\[\Delta U=-|Q|+W\Rightarrow \Delta U+|Q|=W\]