MCQEasyJEE 2023Equation of State of Ideal Gas

JEE Physics 2023 Question with Solution

A bicycle tyre is filled with air at a pressure of 270kPa270 \, \text{kPa} at 27C27^\circ \text{C}. The approximate pressure of the air in the tyre when the temperature increases to 36C36^\circ \text{C} is:

  • A

    270kPa270 \, \text{kPa}

  • B

    262kPa262 \, \text{kPa}

  • C

    278kPa278 \, \text{kPa}

  • D

    360kPa360 \, \text{kPa}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: Initial pressure is P1=270kPaP_1 = 270 \, \text{kPa}, initial temperature is T1=300KT_1 = 300 \, \text{K}, and final temperature is T2=309KT_2 = 309 \, \text{K}.

Find: The final pressure P2P_2 of the air in the tyre.

At constant volume, use Gay-Lussac’s law:

P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2}

Substituting the values shown in the solution:

P2=P1×T2T1=270×309300P_2 = P_1 \times \frac{T_2}{T_1} = 270 \times \frac{309}{300}

Therefore,

P2278kPaP_2 \approx 278 \, \text{kPa}

So the calculated value is 278kPa278 \, \text{kPa}. The solution concludes this value, although it also states 'The Correct Option is B', which disagrees with the listed options. Since 278kPa278 \, \text{kPa} corresponds to option C, the defensible correct option from the working is C.

Common mistakes

  • Using temperature in degree Celsius directly in P1T1=P2T2\frac{P_1}{T_1} = \frac{P_2}{T_2} is incorrect because gas-law ratios require absolute temperature. Convert to Kelvin first.

  • Assuming pressure remains unchanged because the tyre volume is nearly constant is wrong. At constant volume, pressure increases with temperature.

  • Selecting the option label from the page heading without checking the calculation can be misleading here. Verify the numerical result from the working and then match it to the options.

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