Question:
At \(10^{\circ }\text{C}\), the value of density of a fixed mass of a gas (ideal) divided by its pressure is \(x\). At \(110^{\circ }\text{C}\) this ratio is
At \(10^{\circ }\text{C}\), the value of density of a fixed mass of a gas (ideal) divided by its pressure is \(x\). At \(110^{\circ }\text{C}\) this ratio is
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For an ideal gas, density divided by pressure equals \(M/RT\), so it varies inversely with absolute temperature.
Updated On: Nov 12, 2026
- \(x\)
- \(\frac{10x}{283}\)
- \(283x\)
- \(\frac{283x}{383}\)
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The Correct Option is D
Solution and Explanation
Step 1: Understanding the Concept
From \(PV=\dfrac{m}{M}RT\), we get \(\dfrac{\rho}{P}=\dfrac{M}{RT}\), so \(\dfrac{\rho}{P}\propto\dfrac{1}{T}\).
Step 2: Temperatures
\(T_{1}=10+273=283\) K and \(T_{2}=110+273=383\) K.
Step 3: Calculation
\(\dfrac{(\rho/P)_{2}}{(\rho/P)_{1}}=\dfrac{T_{1}}{T_{2}}=\dfrac{283}{383}\)
Step 4: Compare the options
The new ratio is \(\dfrac{283x}{383}\), option (D).
Final Answer:
The new ratio is 283x divided by 383. \[ \boxed{\text{(D) }\dfrac{283x}{383}} \]
From \(PV=\dfrac{m}{M}RT\), we get \(\dfrac{\rho}{P}=\dfrac{M}{RT}\), so \(\dfrac{\rho}{P}\propto\dfrac{1}{T}\).
Step 2: Temperatures
\(T_{1}=10+273=283\) K and \(T_{2}=110+273=383\) K.
Step 3: Calculation
\(\dfrac{(\rho/P)_{2}}{(\rho/P)_{1}}=\dfrac{T_{1}}{T_{2}}=\dfrac{283}{383}\)
Step 4: Compare the options
The new ratio is \(\dfrac{283x}{383}\), option (D).
Final Answer:
The new ratio is 283x divided by 383. \[ \boxed{\text{(D) }\dfrac{283x}{383}} \]
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