SATHEE: Heat and Thermodynamics 5 Question 19

Heat and Thermodynamics 5 Question 19

20. $5.6 L$ of helium gas at STP is adiabatically compressed to $0.7 L$ . Taking the initial temperature to be $T_{1}$ , the work done in the process is

(2011)

(a) $\frac{9}{8} R T_{1}$

(b) $\frac{3}{2} R T_{1}$

(d) $\frac{9}{2} R T_{1}$

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Answer:

Correct Answer: 20. (a)

Solution:

At STP, $22.4 L$ of any gas is 1 mole.

$∴ 5.6 L = \frac{5.6}{22.4} = \frac{1}{4}$ moles $= n$

In adiabatic process, $T V^{γ - 1} =$ constant

$∴ T_{2} V_{2}^{γ - 1} = T_{1} V_{1}^{γ - 1}$ or $T_{2} = T_{1} {(\frac{V_{1}}{V_{2}})}^{γ - 1}$ $γ = \frac{C_{p}}{C_{V}} = \frac{5}{3}$ for monoatomic He gas.

$∴ T_{2} = T_{1} {(\frac{5.6}{0.7})}^{\frac{5}{3} - 1} = 4 T_{1}$

Further in adiabatic process, $Q = 0$

$∴ W + Δ U = 0$

or $W = - Δ U = - n C_{V} Δ T = - n (\frac{R}{γ - 1}) (T_{2} - T_{1})$

$= - \frac{1}{4} (\frac{R}{\frac{5}{3} - 1}) (4 T_{1} - T_{1}) = - \frac{9}{8} R T_{1}$

Heat and Thermodynamics 5 Question 19

20. $5.6 L$ of helium gas at STP is adiabatically compressed to $0.7 L$ . Taking the initial temperature to be $T_{1}$ , the work done in the process is

Answer:

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Heat and Thermodynamics 5 Question 19

20. 5.6L of helium gas at STP is adiabatically compressed to 0.7L. Taking the initial temperature to be T1, the work done in the process is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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20. $5.6 L$ of helium gas at STP is adiabatically compressed to $0.7 L$ . Taking the initial temperature to be $T_{1}$ , the work done in the process is