Q1: A car tyre of volume 0.057 m³ is inflated to 300 kPa at 300 K. After the car is driven for 10 hours, the pressure in the tyre increases to 330 kPa. Assume air is an ideal gas and Cv for air is 21 J/mole.K. The change in the internal energy of air in the tyre in J/mole is

ANS:B - 630

To find the change in internal energy of the air in the tire, we can use the first law of thermodynamics, which states that the change in internal energy (Δ𝑈ΔU) of a system is equal to the heat added to the system (𝑄Q) minus the work done by the system (𝑊W). Mathematically, it can be expressed as: Δ𝑈=𝑄−𝑊ΔU=Q−W Since the tire undergoes an adiabatic process (no heat exchange) and work is done due to the change in pressure, we can use the formula for adiabatic work: 𝑊=−𝑃extΔ𝑉W=−Pext​ΔV Where:

• 𝑃extPext​ is the external pressure (which is the pressure exerted by the atmosphere, assumed to be constant and equal to the final pressure inside the tire),
• Δ𝑉ΔV is the change in volume of the gas.

Given that the tire's volume is initially 0.057 m³ and the pressure increases from 300 kPa to 330 kPa, we can calculate the change in volume using the ideal gas law: 𝑃1𝑉1=𝑛𝑅𝑇1P1​V1​=nRT1​ 𝑃2𝑉2=𝑛𝑅𝑇2P2​V2​=nRT2​ Since the amount of gas (moles) remains constant, we can equate the two expressions: 𝑃1𝑉1=𝑃2𝑉2P1​V1​=P2​V2​ 𝑉2=𝑃1𝑃2×𝑉1V2​=P2​P1​​×V1​ 𝑉2=300 kPa330 kPa×0.057 m3V2​=330kPa300kPa​×0.057m3 𝑉2≈0.0518 m3V2​≈0.0518m3 Now, we can calculate the work done: 𝑊=−𝑃extΔ𝑉W=−Pext​ΔV 𝑊=−(330×103 Pa)×(0.0518−0.057) m3W=−(330×103Pa)×(0.0518−0.057)m3 𝑊≈−3090 JW≈−3090J Since this is work done by the gas, we use the negative sign. Now, we can calculate the change in internal energy (Δ𝑈ΔU) using the first law of thermodynamics: Δ𝑈=𝑄−𝑊ΔU=Q−W Since the process is adiabatic (𝑄=0Q=0), we have: Δ𝑈=−𝑊ΔU=−W Δ𝑈≈3090 JΔU≈3090J Therefore, the change in internal energy of air in the tire is approximately 3090 J.