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1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from        25°C to 100°C. The DH(cal) is:
    Correct answer is '375'. Can you explain this answer?
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    1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a ...
    °C to 50°C.

    We can use the formula for the heat transfer at constant pressure, which is:

    Q = nCpΔT

    where Q is the heat transferred, n is the number of moles of gas, Cp is the specific heat capacity at constant pressure, and ΔT is the change in temperature.

    Since the gas is ideal and CV = 3/2 R, we can use the relationship between Cp and CV:

    Cp = CV + R

    Therefore, Cp = 5/2 R for this gas.

    We are given that n = 1 mole, Cp = 5/2 R, and ΔT = 50°C - 25°C = 25°C = 298 K.

    Plugging in these values, we get:

    Q = nCpΔT = (1 mol)(5/2 R)(298 K) = 373.75 J

    Therefore, the heat transferred to the gas is 373.75 J.
    Community Answer
    1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a ...
    If we take R = 2 then we get 375 . if we take acurate value of R then we get 372.93.
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    1 mole of an ideal gas for which CV = 3/2 R is heated reversibly at a constant pressure of 1 atm from 25°C to 100°C. The DH(cal) is:Correct answer is '375'. Can you explain this answer?
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