Question

Question: In an isobaric compression of an ideal gas,

(a) no heat flows into the gas.

(b) the internal energy of the gas remains constant.

(c) no work is done on the gas.

(d) work is done on the gas.

(e) work is done by the gas.

Short answer

In isobaric compression of an ideal gas, work is done on the gas.

Step-by-step solution

Understanding isobaric process

An isobaric process is a thermodynamic process in which the pressure stays constant.

Consider a gas is compressed isobaricallyata constant pressure P.

The expression for heat transfer in this process is

\[Q = n{C_{\rm{P}}}\Delta T\].

Here, \[n\] is the number of moles, \[{C_{\rm{P}}}\]is the specific heat at constant pressure, and \[\Delta T\] is the difference in temperature.

The expression for the work done in an isobaric process is

\[W = P\Delta V\].

Here, \[\Delta V\]is the change in volume.

Explanation of the given options 

In an isobaric process, the temperature increases. So,\[\Delta T\]and\[Q\]are positive. The heat flows into the gas.

Thus, option (a) is incorrect.

In an isobaric process, the heat transferred to the system does work. But it also changes the internal energy of the system.

Thus, option (b) is incorrect.

Since the change in volume is not zero, work is also not zero.

Thus, option (c) is incorrect.

The sign of \[\Delta V\] determines whether the work is done on the gas or by the gas. For the compression process, the volume decreases. Therefore, \[\Delta V\] and the work done are negative. It means work is done on the gas.

Thus, option (e) is incorrect, and option (d) is correct.