Tag: principal and molar specific heats of gases

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Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A gas expands against a constant external pressure of  $2.00 atm, $ increasing its volume by $ 3.40 L.$   Simultaneously, the system absorbs  $400 J $ of heat from its surroundings. What is  $ \Delta E ,$  in joules, for this gas?

  1. $- 689$

  2. $-289$

  3. $+400$

  4. $+289$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the first law of thermodynamics, delta E = Q - W. Work done by the gas is P * delta V = 2.00 atm * 3.40 L = 6.80 L*atm. Converting to Joules (1 L*atm = 101.3 J), W = 6.80 * 101.3 = 688.84 J. Thus, delta E = 400 J - 688.84 J = -288.84 J, which rounds to -289 J.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Consider a classroom that is roughly  $5 { m } \times 10  { m } \times 3  { m }.$  Initially   ${ t } = 20 ^ { \circ }  { C }$  and  $ { P } = 1$ atm. There are  $50$  people in an insulated class loosing energy to the room at the average rate of  $150$  watt per person. How long can they remain in class if the body temperature is  $37 ^ { \circ } \mathrm { C }$  and person feels uncomfortable above this temperature. Molar heat capacity of air  $= ( 7 / 2 ) R.$

  1. $4.34$ minutes

  2. $5.73$ minutes

  3. $6.86$ minutes

  4. $7.79$ minutes
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The room volume is 150 m^3. Using PV=nRT, calculate the number of moles of air. The total heat added by 50 people is 50 * 150 W = 7500 J/s. The heat required to raise the air temperature from 20 C to 37 C is Q = n * Cv * delta T. Solving for time t = Q / Power gives approximately 4.34 minutes.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Assertion : $C _P$ is always greater than $C _V$ in gases.
Reason : Work done at constant pressure is more than at constant volume.

  1. If both assertion and reason are true and reason is the correct explanation of assertion

  2. If both assertion and reason are true but reason is not the correct explanation of assertion

  3. If assertion is true but reason is false

  4. If both assertion and reason are false
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Cp is greater than Cv because at constant pressure, the system does work on the surroundings as it expands, requiring additional heat input to achieve the same temperature rise. The reason correctly explains the assertion.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

$C _{P}$ and $C _{V}$ are specific heats at constant pressure and constant volume, respectively. It is observed that $C _{P} - C _{V} = a$ for hydrogen gas $C _{P} - C _{V} = b$ for nitrogen gas. The correct relation between $a$ and $b$ is

  1. $a = b$

  2. $a = 14b$

  3. $a = 28b$

  4. $a = \dfrac {1}{14}b$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For any ideal gas, Cp - Cv = R (in molar terms). If 'a' and 'b' refer to specific heats (per unit mass), then Cp - Cv = R/M, where M is molar mass. For hydrogen, M=2; for nitrogen, M=28. Thus, a = R/2 and b = R/28. Therefore, a/b = 28/2 = 14, so a = 14b.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

If $C _{p} and C _{v}$ denoto the specific heats of nitron per unit mass at constant pressure and constant volume rest then 

  1. $C _{p} and C _{v}$=R/28

  2. $C _{p} and C _{v}$=R/14

  3. $C _{p} and C _{v}$=R

  4. $C _{p} and C _{v}$=28R
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

According to Mayer's relation $C _{p}-C _{v}= R/m$

$C _{p}-C _{v}=\dfrac{R}{m}$
for nitrogen $m=28$
$ \therefore C _{p}- C _{v^{2}} R/28$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Ration of $C _p$ and $C _v$ depends upon temperatures according to the following relation

  1. $\gamma \propto T$

  2. $\displaystyle \gamma \propto \frac{1}{T}$

  3. $\gamma \propto \sqrt{T}$

  4. $\gamma \propto T^o$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\gamma =\dfrac{C _p}{C _v}$ i.e, ratio of specific heat capacity at constant pressure and specific heat capacity at constant volume. It doesn't depend on temperature, i.e, it is independent of temperature.