Q117 / 150 · MHT CET 2026 (18 April Shift 2) · Physics
At a pressure P and temperature T, 7 gram of oxygen occupies a volume V. The equation of state will be (Molecular weight of Oxygen $= 32$)
- A$PV = \left(\dfrac{2}{7}\right)RT$
- B$PV = \left(\dfrac{7}{2}\right)RT$
- C$PV = \left(\dfrac{7}{16}\right)RT$
- Correct answerD$PV = \left(\dfrac{7}{32}\right)RT$
Solution
The ideal gas equation is given by $PV = nRT$.
The number of moles $n$ is calculated as $n = \dfrac{m}{M}$.
Given mass of oxygen $m = 7$ g and molecular weight $M = 32$ g/mol.
Therefore, $n = \dfrac{7}{32}$.
Substituting the value of $n$ into the ideal gas equation, we get $PV = \left(\dfrac{7}{32}\right)RT$.
The number of moles $n$ is calculated as $n = \dfrac{m}{M}$.
Given mass of oxygen $m = 7$ g and molecular weight $M = 32$ g/mol.
Therefore, $n = \dfrac{7}{32}$.
Substituting the value of $n$ into the ideal gas equation, we get $PV = \left(\dfrac{7}{32}\right)RT$.
Hence, the correct answer is option: 'D'.