SATHEE: Thermodynamics Question 327

Thermodynamics Question 327

Question: A 100 kg piston encloses 32 g of oxygen gas at a temperature of $27^{o} C$ in a vertical cylinder of base area of $4 d m^{2}$ . The air pressure outside is $1 \times 10^{5} P a$ . The axis of the cylinder is vertical, and the piston can move in it without friction. How much heat is to be transferred to the gas to raise the piston by 20 cm. Use $R = \frac{25}{3} J / m o l / K$

Options:

A) 3500 J

B) 350 J

C) 7000 J

D) 750 J

Show Answer

Answer:

Correct Answer: A

Solution:

[a] As heat is supplied to gas, it expands against atmospheric pressure and weight of piston. Thus the pressure of gas is constant given by,

$P = P_{0} + \frac{m g}{A}$

$P = 1 \times 10^{5} + \frac{100 \times 10}{4 \times 10^{- 2}} = (1 + \frac{1}{4}) 10^{5}$

$= 1.25 \times 10^{5} N / m^{2}$

It is given that,

$A = 4 d m^{2} = 4 \times 10^{- 2} m^{2}$

$m = 100 k g$

$n = 1mole$

$V = \frac{n R T}{P} = 2 \times 10^{- 2} m^{3}$

Initial height of piston from base of vessel,

$h = \frac{V}{A} = \frac{2 \times 10^{- 2}}{4 \times 10^{- 2}} = \frac{1}{2} m$

As process is isobaric, so

$\frac{T_{1}}{T_{2}} = \frac{V_{1}}{V_{2}} = \frac{V}{V + A Δ h} = \frac{h}{h + Δ h}$

$∴$ $\frac{T_{2} - T_{1}}{T_{1}} = \frac{Δ h}{h}$

$Δ Q = n C_{p} (T_{2} - T_{1}) = \frac{n . C_{p} . T_{1}}{h} Δ h$

$= \frac{7}{2} \times \frac{25}{3} \times \frac{300}{0.5} \times 0.2 = 350 0 J$

Thermodynamics Question 327

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Thermodynamics Question 327

Options:

Answer:

Solution:

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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