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NEET
Exam
NEET
Subject
Physics
Chapter
Kinetic Theory of Gases

NEET · Physics · Kinetic Theory of Gases

The pressure ( P ) versus temperature ( T in Kelvin) graph for a...

The pressure (PP) versus temperature (TT in Kelvin) graph for a fixed mass of an ideal gas shows two straight lines passing through the origin, corresponding to two different constant volumes V1V_1 and V2V_2. The line for V1V_1 is steeper than the line for V2V_2. Two states of the gas, AA and BB, are marked on the lines for V1V_1 and V2V_2 respectively, such that both states have the exact same pressure P0P_0. If vAv_A and vBv_B are the root mean square (rms) speeds of the gas molecules in states AA and BB respectively, which of the following is correct?

Options

  1. AvA<vBv_A < v_B
  2. BvA>vBv_A > v_B
  3. CvA=vBv_A = v_B
  4. DvA=2vBv_A = 2v_B

Correct answer

A. vA<vBv_A < v_B

Step-by-step solution

From the ideal gas equation, PV=nRT⇒P=(nRV)TPV = nRT \Rightarrow P = \left(\dfrac{nR}{V}\right)T.

For a P−TP-T graph, the slope is nRV\dfrac{nR}{V}. Since the line for V1V_1 is steeper than the line for V2V_2, its slope is greater, which means V1<V2V_1 < V_2.

States AA and BB are at the same pressure P0P_0. From the graph equation T=PVnRT = \dfrac{PV}{nR}, at a constant pressure P0P_0, the temperature is directly proportional to the volume. Since V1<V2V_1 < V_2, the temperature of state AA (on V1V_1) is less than the temperature of state BB (on V2V_2), so TA<TBT_A < T_B.

The rms speed of gas molecules is given by vrms=3RTMv_{rms} = \sqrt{\dfrac{3RT}{M}}. Since vrms∝Tv_{rms} \propto \sqrt{T} and TA<TBT_A < T_B, it follows that vA<vBv_A < v_B.

Answer: vA<vBv_A < v_B
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