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Exercises · 11.2

Q.What amount of heat must be supplied to 2.0×10−2 kg2.0 \times 10^{-2}\ \text{kg} of nitrogen (at room temperature) to raise its temperature by 45 ∘C45\,^{\circ}\text{C} at constant pressure? (Molecular mass of N2=28\text{N}_2 = 28; R=8.3 J mol−1 K−1R = 8.3\ \text{J mol}^{-1}\ \text{K}^{-1}.)

Rajasthan RbseTextbookSubjective· 3mImportance★★★★★est
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✓ Free question

For an ideal gas at constant pressure, heat supplied is Q=nCpΔTQ = n C_p \Delta T. Using Cp=72RC_p = \frac{7}{2}R for diatomic nitrogen, the heat required is 933 J\boxed{933\,\text{J}}.

When we heat a gas at constant pressure, it not only gains internal energy but also does work by expanding against the external pressure. This is why the heat capacity at constant pressure, CpC_p, is always larger than at constant volume, CvC_v. For an ideal gas, the relationship is Cp=Cv+RC_p = C_v + R.

Nitrogen is a diatomic molecule. At room temperature, it has three translational and two rotational degrees of freedom (vibrational modes are not excited). By the equipartition theorem, each degree of freedom contributes 12R\frac{1}{2}R per mole to the molar heat capacity at constant volume:

Cv=52RC_v = \frac{5}{2}R

Therefore, the molar heat capacity at constant pressure is:

Cp=Cv+R=52R+R=72RC_p = C_v + R = \frac{5}{2}R + R = \frac{7}{2}R

Cp=72R=72×8.3=29.05 J mol−1K−1C_p = \frac{7}{2}R = \frac{7}{2} \times 8.3 = 29.05\,\text{J mol}^{-1}\text{K}^{-1}

Now let's calculate the heat required step by step:

  1. Find the number of moles of nitrogen.

    Given mass m=2.0×10−2 kg=20 gm = 2.0 \times 10^{-2}\,\text{kg} = 20\,\text{g} and molecular mass M=28 g mol−1M = 28\,\text{g mol}^{-1}:

n=mM=2028=57 moln = \frac{m}{M} = \frac{20}{28} = \frac{5}{7}\,\text{mol}

  1. Identify the temperature change.

    The temperature rise is ΔT=45 ∘C=45 K\Delta T = 45\,^{\circ}\text{C} = 45\,\text{K} (since a change in Celsius equals a change in Kelvin).

  2. Apply the heat capacity formula at constant pressure.

    The heat supplied at constant pressure is:

Q=nCpΔTQ = n C_p \Delta T

Substituting the values:

Q=57×72×8.3×45Q = \frac{5}{7} \times \frac{7}{2} \times 8.3 \times 45

  1. Simplify the calculation.

    Notice that 57×72=52\frac{5}{7} \times \frac{7}{2} = \frac{5}{2}:

Q=52×8.3×45=2.5×8.3×45Q = \frac{5}{2} \times 8.3 \times 45 = 2.5 \times 8.3 \times 45

Q=2.5×373.5=933.75 JQ = 2.5 \times 373.5 = 933.75\,\text{J}

Watch out

A common mistake is to use CvC_v instead of CpC_p when the problem specifies constant pressure. Always check whether the process is isobaric (constant PP) or isochoric (constant VV).

✓Final answer

The amount of heat that must be supplied is 933 J\boxed{933\,\text{J}} (or 934 J934\,\text{J} if rounded).

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