MHT CET · Physics · Thermodynamics
For an ideal gas, the density of the gas is \(\varrho_0\) when temperature and pressure of the gas are \(T_0\) and \(P_0\) respectively. when the temperature of the gas is \(2 \mathrm{~T}_0\), its pressure becomes \(3 \mathrm{P}_0\). The new density will be
- A \(\frac{2}{3} \varrho_0\)
- B \(\frac{3}{4} \varrho_0\)
- C \(\frac{4}{3} \varrho_0\)
- D \(\frac{3}{2} \varrho_0\)
Answer & Solution
Correct Answer
(D) \(\frac{3}{2} \varrho_0\)
Step-by-step Solution
Detailed explanation
\(\varrho = \frac{PM}{RT}\) \(\varrho_0 = \frac{P_0M}{RT_0}\)
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