NEET
Practice Test 9
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A mass of ideal gas at pressure P is expanded isothermally to four times the original volume and then slowly compressed adiabatically to its original volume. Assuming γ to be 1.5, the new pressure of the gas is

  • Solution

    Let P and V be the initial pressure and volume of ideal gas. After isothermal expansion, pressure is P/4. So volume is 4 V.
    Let P1 be the pressure after adiabatic compression.
    Then
    P1Vγ=(P / 4)(4 V)γ
    P1=(P / 4)(4)3/2=2 P

A system changes from the state (P1, V1) to (P2, V2) as shown in the figure. What is the work done by the system?

  • Solution

    \(W=\frac{(1\times 10^{5}+6\times 10^{5}) (5-1)}{2}\)

    \(=\frac{6\times 10^{5}\times 4}{2}=12\times 10^{5}\) joule

A refrigerator works between 0°C and 27°C. Heat is to be removed from the refrigerated space at the rate of 50 k cal/minute, the power of the motor of the refrigerator is

  • Solution

    \(\frac{T_{2}}{T_{1}-T_{2}}=\frac{Q_{2}}{W}\)

    \(\frac{273}{300-273}=\frac{50,000}{W}\)

    \(W=\frac{27\times 50,000}{273}\)cal/min

    \(p=\frac{W}{t}=\frac{4.2\times 27\times 50,000}{60\times 273}\)Joule/sec

    = 346 watt = 0.346 kW

A perfect gas goes from a state A to another state B by absorbing 8 × 105 J of heat and doing 6.5 × 105 J of external work. It is now transferred between the same two states in another process in which it absorbs 105 J of heat. In these condprocess

  • Solution

    dU = dQ - dW = (8 × 105 - 6.5 × 105) = 1.5 × 105J

    dW = dQ - dU = 105 - 1.5 × 105 = -0.5 × 105J – vesign indicates that work done on the gas is 0.5 × 105J.

The temperature of 5 moles of a gas which was held at constant volume was changed from 100° to 120°C. The change in the internal energy of the gas was found to be 80 joule,the total heat capacity of the gas at constant volume will be equal to

  • Solution

    dU=nCvdt or 80=5×Cv(120-100)
    Cv= 4.0 joule/K

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