TS EAMCET · Physics · Thermodynamics
An ideal gas in a cylinder is compressed adiabatically to one-third of its original volume. A work of \(45 \mathrm{~J}\) is done on the gas by the process. The change in internal energy of the gas and the heat flowed into the gas, respectively are
- A \(45 \mathrm{~J}\) and zero
- B \(-45 \mathrm{~J}\) and zero
- C \(45 \mathrm{~J}\) and heat flows out of the gas
- D \(-45 \mathrm{~J}\) and heat flows into the gas
Answer & Solution
Correct Answer
(A) \(45 \mathrm{~J}\) and zero
Step-by-step Solution
Detailed explanation
For adiabatic process, \(\Delta Q=0\) Now, from first law of thermodynamics, we have \[ \Delta Q=\Delta U+\Delta W \] \(\Delta W=-45\) or work is done on the gas. \[ \begin{array}{lc} \Rightarrow & 0=\Delta U-45 \\ \Rightarrow & \Delta U=45 \mathrm{~J} \end{array} \]
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