Physics

Thermodynamics and Gas Laws

626 Questions

Thermodynamics and gas laws questions test the understanding of ideal gas behavior, work done during thermodynamic processes, and specific heat ratios. Key areas include isothermal, adiabatic, and isobaric expansions along with real gas deviations. These mathematical physics concepts are standard in engineering and general science competitive exams.

Ideal gas equationIsothermal and adiabatic processesThermodynamic workGas kinetic theoryReal gas behavior

Thermodynamics and Gas Laws Questions

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

The state of gas can be described by quoting the relationship between_____________.

  1. pressure, volume, temperature

  2. temperature, amount, pressure

  3. amount, volume, temperature

  4. pressure, volume, temperature, amount
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The pressure $(P)$, volume $(V)$, temperature $(T)$, amount $(n)$ etc. are the state variables or state functions.

Multiple choice chemistry chemical thermodynamics system and surroundings introduction to thermodynamics basics of thermodynamics

A container of volume $1{m}^{3}$ is divided into two equal parts by a partition. One part has an ideal diatomic gas at $300K$ and the other part has vacuum. The whole system is isolated from the surrounding. When the partition is removed, the gas expands to occupy the whole volume. Its temperature will be:

  1. $300K$
  2. ${ 227.5 }^{ o }C$
  3. $455K$
  4. ${455}^{o}C$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a free expansion of an ideal gas into a vacuum. Since the system is isolated (q=0) and expands into a vacuum (w=0), the internal energy remains constant (delta U = 0). For an ideal gas, internal energy depends only on temperature, so the temperature remains 300K.

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

The amount of heat necessary to raise the temperature of $0.2 \ mol\ of\ N _2$ at constant pressure from $37^oC$ to $ 337^oC$  will be

  1. $746\ J$
  2. $1746\ J$
  3. $2746\ cal$
  4. $3746\ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$N _2$ is a diatomic molecule thus its degree of freedom is 5. Its $C _p$ is given as $(1+\displaystyle\dfrac{f}{2})R=(1+\dfrac{5}{2})R=\dfrac{7}{2}R$
Thus, we get the heat required as $Q=nC _p\Delta T=0.2\times \displaystyle\dfrac{7}{2}\times 8.314\times 300=1746  J$

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

The specific heat of a gas at constant pressure as compared to that at constant volume is

  1. less

  2. equal

  3. more

  4. constant
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

When the gas is heated at constant pressure, some amount of heat is used up in increasing the volume of the gas. For a constant volume process no such heat is required. Thus $C _p>C _v$

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

The molar specific heat of an ideal gas at constant pressure and volume are $C _p$ and $C _v$ respectively. The value of $C _v$ is

  1. $R$
  2. $\gamma$ R
  3. $\dfrac{R}{\gamma-1}$
  4. $\dfrac{\gamma R}{\gamma-1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We know that $\displaystyle\dfrac{C _p}{C _v}=\gamma$ and $C _p-C _v=R$.
Thus we get $C _v(\displaystyle\dfrac{C _p}{C _v}-1)=R$
or, $C _v=\displaystyle\dfrac{R}{\gamma -1}$

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

Calculate the specific heat of a gas at constant volume from the following data. Density of the gas at N.T.P =$19 \times 10 ^ { - 2 } \mathrm { kg } / \mathrm { m } ^ { 3 }$ $\left( C _ { p } / C _ { v } \right)$ = 1.4,J =$4.2 \times 10 ^ { 3 } \mathrm { J } / \mathrm { kcal }$ atmospheric pressure=$1.013 \times 10 ^ { 5 } N / m ^ { 2 }$ (in kcal /kg k)

  1. $2.162$
  2. $1.612$
  3. $1.192$
  4. $2.612$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the relationship Cp - Cv = R/M and Cp/Cv = gamma, we can solve for Cv using the density and pressure at NTP. The calculation yields approximately 2.162 kcal/kg K.

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

The ratio of the specific heat of air at constant pressure to its specific heat constant volume is

  1. Zero

  2. Greater than one

  3. Less than one

  4. Equal to one
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The correct answer is option(B).

The ratio of specific heat at constant pressure to the specific heat at constant volume is always greater than one.
As, when the gas is allowed to expand resulting in constant pressure, some of the heat is converted to work resulting in the need of a higher amount of heat to raise the temperature of the gas. Whereas when the volume of the gas is constant, the entire heat supplied is utilized in raising the gas temperature. Hence the heat required the raise the temperature of a unit mass of gas at constant pressure is greater than that required at constant volume. Hence the ratio $c _p:c _v$ is always greater than one.

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

For a gas the ratio of the two specific heats is $\dfrac{5}{3}$. If R $=$ 2 cal /mol-K then the values of $C _{p}$ and $C _{v}$ in cal / mol- K 

  1. $C _p=5 ,C _v=3 $
  2. $C _p=3 ,C _v=4 $
  3. $C _p=4 ,C _v=3 $
  4. $C _p=3 ,C _v=5 $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From given data we have $C _{p} - C _{v} = 2 $
and $\dfrac{C _{p}} { C _{v}} = \dfrac {5}{3} $
Solving both gives , 
$C _{p} = 5$ and  $ C _{v} = 3 $

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

A diatomic gas molecule has translational, rotational and vibrational degrees of freedom. Then $\dfrac{C _{p}}{C _{v}}$ is

  1. 1.67

  2. 2.14

  3. 1.29

  4. 1.33
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The diatomic molecule has total 6 degress of freedom(3 translational, 2 rotational and 1 vibrational)
Now $C _p$ is given as $(1+\dfrac{f}{2})R=(1+\dfrac{6}{2})R=4R$
and $C _v$ is given as $\dfrac{f}{2}R=\dfrac{6}{2}R=3R$
Thus we get $\dfrac{C _p}{C _v}=\dfrac{4}{3}=1.33$

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

If the ratio of sp.heat of a gas at constant pressure to that at constant volume is $\gamma $ , the change in internal energy of gas, when the volume changes from V to 2V at constant pressure P is 

  1. $\dfrac{R}{\gamma -1}$

  2. PV
  3. $\dfrac{PV}{\gamma -1}$
  4. $\dfrac{\gamma PV}{\gamma -1}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The change in internal energy in the process should have been,
U = $ nC _v \Delta T $
Now, for this process, if $ \dfrac{{C} _{p}}{{C} _{v}} = \gamma $ and $C _p-C _v=R$
Then, $C _v =  \dfrac{R}{\gamma - 1} $
U = $ \dfrac{nR \Delta T}{\gamma - 1} $
Now, $ nR \Delta T = P(2V - V) $
Thus, U = $ \dfrac{PV}{\gamma - 1} $

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

A cylinder of fixed capacity $67.2$ liters contains helium gas at STP. Calculate the amount of heat required to raise the temperature of the gas by $15^{o}C$. ($R=8.314\ J\ mol^{-1}k^{-1}$)

  1. $520\ J$
  2. $560. J$
  3. $620\ J$
  4. $621.2\ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since, the process is at constant volume,
Q = U as W = 0
Thus, Q = $ n {C} _{v} \Delta T $
At STP, n = $ \dfrac{PV}{RT} $
Since, He is diatomic, $ {C} _{v} = 2.5R $
Q = $ \dfrac{PV}{RT} \times 2.5R \times 15 $
Substituting the pressure and temperature values at STP, 
P = 1 atm
V = 67.2 L
T = 298 K
we get,
Q = 560.9 J

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

A diatomic gas is heated at constant pressure. The fraction of the heat energy used to increase the internal energy is 

  1. $ \dfrac{3}{5}$
  2. $ \dfrac{3}{7}$
  3. $ \dfrac {5}{7}$
  4. $ \dfrac {7}{9}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In a diatomic gas, we have $C _p=\dfrac{7}{2}R $ and $C _v=\dfrac{5}{2}R$
The heat is given as $nC _p\Delta T$ and internal energy as $nC _v\Delta T$
Thus we get $\dfrac{U}{Q}$ as $\dfrac{5}{7}$

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

Four students found set of $C _{p}$ and $C _{v}$[in cal/deg mole] as given below, which of the following set is correct 

  1. $C _{v}=4,C _{p}=2$
  2. $C _{v}=4,C _{p}=3$
  3. $C _{v}=3,C _{p}=4$
  4. $C _{p}=5,C _{v}=3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

C$ _{v}$ cannot be greater than $C _{p}$

Hence, option A  and option B are incorrect.

We have the relation $C _{p} - C _{v}$= R ( and its value is 2 cal/mole ) and hence option C is also incorrect.

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

If $C _p$ and $C _v$ denote the specific heats (per unit mass) of an ideal gas of molecular weight M, where R is the molar gas constant:

  1. $C _p - C _v = R/M^2$
  2. $C _p - C _v = R$
  3. $C _p - C _v = R/M$
  4. $C _p - C _v = M/R$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

By definition 

$dU={ C } _{ v }dT\longrightarrow 1$

also enthalpy,

$H=U+PV\\ or\quad dH=dU+d\left( PV \right) \\ or\quad dH=dU+nRdT\longrightarrow 2$

Also $dH={ C } _{ P }dT\\ \therefore { C } _{ P }dT={ C } _{ V }dT+nRdT\\ \Rightarrow { C } _{ P }={ C } _{ V }+nR\\ or{ C } _{ P }-{ C } _{ V }\quad =nR=\cfrac { Rm }{ M } $

for $m=1$

${ C } _{ P }-{ C } _{ V }=\cfrac { R }{ M } $