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Heat and Thermodynamics question

2012 · Shift 2 · Q45
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Heat and Thermodynamics question

2012 · Shift 2 · Q45

JEE AdvancedPhysicsHeat and ThermodynamicsMCQ+3 / −0.75
Two moles of ideal helium gas are in a rubber balloon at 30oC. The balloon is fully expandable and can be assumed to require no energy in its expansion. The temperature of the gas in the balloon is slowly changed to 35oC. The amount of heat required in raising the temperature is nearly (take R = 8.31 J/mol.K)
  1. A
    62 J
  2. B
    104 J
  3. C
    124 J
  4. D
    208 J
View written solutionFree

Correct answer: D

Step-by-step Solution

  1. Identify the Given Information

    • Number of moles of helium gas, n = 2 mol.
    • Initial temperature, T1=30∘CT_1 = 30^\circ CT1​=30∘C.
    • Final temperature, T2=35∘CT_2 = 35^\circ CT2​=35∘C.
    • The gas is Helium, which is a monatomic ideal gas.
    • The balloon is fully expandable and requires no energy for expansion. This implies the process occurs at constant pressure (isobaric process), where the internal pressure is always equal to the constant external atmospheric pressure.
    • The universal gas constant, R = 8.31 J/mol.K.

  2. Calculate the Change in Temperature (ΔT\\\Delta TΔT) The change in temperature can be calculated in Celsius or Kelvin, as the difference is the same. ΔT=T2−T1=35∘C−30∘C=5∘C=5 K\Delta T = T_2 - T_1 = 35^\circ C - 30^\circ C = 5^\circ C = 5 \text{ K}ΔT=T2​−T1​=35∘C−30∘C=5∘C=5 K

  3. Determine the Molar Specific Heat at Constant Pressure (CpC_pCp​) For a thermodynamic process at constant pressure, the heat required is given by Q=nCpΔTQ = n C_p \Delta TQ=nCp​ΔT.

    • Helium (He) is a monatomic gas. For a monatomic ideal gas, the molar specific heat at constant volume (CvC_vCv​) is given by: Cv=32RC_v = \frac{3}{2}RCv​=23​R
    • The molar specific heat at constant pressure (CpC_pCp​) is related to CvC_vCv​ by Mayer's relation: Cp=Cv+RC_p = C_v + RCp​=Cv​+R
    • Substituting the expression for CvC_vCv​ for a monatomic gas: Cp=32R+R=52RC_p = \frac{3}{2}R + R = \frac{5}{2}RCp​=23​R+R=25​R

  4. Calculate the Amount of Heat Required (Q) Now, we can substitute the values into the formula for heat supplied in an isobaric process: Q=nCpΔTQ = n C_p \Delta TQ=nCp​ΔT Q=n(52R)ΔTQ = n \left( \frac{5}{2}R \right) \Delta TQ=n(25​R)ΔT Plugging in the given values: Q=(2 mol)×(52×8.31Jmol⋅K)×(5 K)Q = (2 \text{ mol}) \times \left( \frac{5}{2} \times 8.31 \frac{\text{J}}{\text{mol}\cdot\text{K}} \right) \times (5 \text{ K})Q=(2 mol)×(25​×8.31mol⋅KJ​)×(5 K) Q=2×52×8.31×5 JQ = 2 \times \frac{5}{2} \times 8.31 \times 5 \text{ J}Q=2×25​×8.31×5 J Q=5×8.31×5 JQ = 5 \times 8.31 \times 5 \text{ J}Q=5×8.31×5 J Q=25×8.31 JQ = 25 \times 8.31 \text{ J}Q=25×8.31 J Q=207.75 JQ = 207.75 \text{ J}Q=207.75 J

  5. Compare with the Options The calculated amount of heat is 207.75 J. This value is approximately 208 J.

    • A: 62 J
    • B: 104 J
    • C: 124 J
    • D: 208 J

    The closest option is D.

Conclusion

The amount of heat required in raising the temperature is nearly 208 J.

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