21 + Interview Questions in SPECIFIC-HEAT-CAPACITIES-OF-GASES IN GENERAL AWARENESS Page 1 InterviewSolution

1.	A sample of air weighing `1.18 g` occupies `1.0 xx 10^(3) cm^(3)` when kept at `300 K and 1.0 xx 10^(5)` pa. When `2.0 cal` of heat is added to it constant volume, its temperature increases by `1^@C`. Calculate the amount if heat needed to increases the temperature of air by `1^@C` at constant pressure if the mechanical equivalent of heat si `4.2 xx 10^(-1)` . Assume that air behaves as an ideal gas.
Answer» `m=1.18g` `V=1xx10^(3) cm^(3)= 1L` `T=300K`, `P=10^(5) Pa` `PV =nRT` or `n=(PV)/(RT)` `[10^(5)Pa=1atm.]` implies `n=(1)/(8.2xx10^(-2)xx3xx10^(2)` implies `=(1)/(8.2xx3)=1/24.6` Now, `C_v =1/nxx2 = 24.6 xx 2 =49.2` `C_p =R+C_v = 1.987 +49.2` `=51.187` `Q=nC_p dT` `=(1)/(24.6)xx 51.187xx1` `=2.08 cal`.

Discussion

2.	Figure shows a cylindrical container containing oxyegn `(gamma = 1.4)` and closed by a 50 kg frictionless piston. The area of cross section is `100cm(2)` , atmospheric pressure is `100 kPa`+E2 and g is 10 m s^(-2)`. The cylinder is slowly heated for some time. Find the amount of heat supplied to gas if the piston moves out through a distsnce of 20 cm. .
Answer» `gamma =1.4`, Weight of piston (w) `= 50 kg`, Area of cross section of piston(A) `=100 cm^2` Atmospheric pressure `(P_0) = 100 kPa`, `g=10 m//s^2, x=20 cm`, `dW = Pdv= ((mg)/A+P_0).Adx` `= ((50xx10)/(100xx10^(-4)) + 10^(5))` `(100 xx 10 ^(-4) xx 20 xx 10^(-2))` `= (5xx10^(4) xx10^(5) ) xx 20 xx 10^(-4)` `= 1.5 xx 10^(5) xx 20 xx 10^(-4)` `=300 J`. Hence `nRdT = 300`ltbgt `dT= 300/nR` ltbgt So, `dQ=nCpdT=nc_p xx((300)/(nR))` `=(ngammaR300)/((gamma-1)nR` `= ((300xx1.4)/0.4)` `=1050J`

Discussion

3.	The volume of an ideal gas `(gamma 1.5 )` is changed adiabatically from 4.00 liters to 3.00 liters . Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.A.B.C.D.
Answer» `gamma=1.5` Since it is an adiabatic process, So, `PV^(gamma) =constant`. (a) `P_1V_1^(gamma) = P_2V_2^(gamma)` Given, `V_1 = 4L`, `V_2 = 3L`, `(P_2)/(P_1)=?` implies`P_2/P_1=((V_1)/(V_2))^gamma` `=(4/3)^(15)=1.5396=1.54` (b) `TV^(gamma-1) =Constant` `T_1V_1^(gamma-1) = T_2V_2^(gamma-1)` implies ` T_2/T_1=((V_1)/(V_2))^(gamma-1)=((4)/(3))^(0.5)=1.154`.

Discussion

4.	An ideal gas at pressure` 2.5 xx 10^(5)` `pa` and temperture `300k` occupies `100 cc`. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure, (b) the final temperature and ( c) the work done by the gas in the peocess. Take `(gamma = 1.5)`.
Answer» `P_1=2.5xx10^(5) Pa`, `T_1=300K` `V_1 =100CC`, (a) `P_1V_1^(gamma)=P_2V_2^(gamma)` implies `2.5 xx 10^5 xx V^1.5=((V)/(2))^1.5 xx P_2` implies `P_2=7.07xx10^5` `=7.1xx10^5` (b) `T_1V_1^(gamma-1)=T_2V_2^(gamma-1)` implies`300xx(100)^(1.5-1)=T_2xx ((100)/(2))^(1.5-1)` `=T_2xx(50)^(1.5-1)` implies ` 300 xx 10 = T_2 xx 7.07` `T_2 = 424.32 K=424K`. (c) Work done by the gas in the process `W=(mR)/(gamma-1)[T_2-T_1]` `=(P_1V_1)/(gamma-1)[T_2-T_1]` `=(2.5xx10)/(300xx0.5)xx124` `=20.67=21J`.

Discussion

5.	Air `(gamma = 1.4 )` is pumped at 20atm pressure in a motor tyre at `20^@C`. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmoshpheric air.
Answer» `gamma=1.4`. `T_1 = 20^@C=293K`, `P_1=2atm`,`P_2=1atm` We know for adiabatic process `P_1^(1-gamma) xx T_1^(1-gamma)=P^(1-gamma) xx T_2^gamma` or `(2)^(1-1.4) xx (293)^1.4 = (1)^(1-1.4) xx T_2^(1.4)` implies `(2)^(-0.4) xx (293)^1.4 = T_2^1.4` implies `2153.78 = T_2^1.4` `T_2=(2153.78)^(1//1.4)` `=240.3K`.

Discussion

6.	A given sample of an ideal gas `(gamma = 1.5 )` is compressed adiabatically from a volume of `150 cm ^(3)` to `50 cm ^(3)`. The initial pressure and the initial temperature are `150 kpa` and `300K`. Find (a) the number of moles of the gas in the sample, (b) the molar heat capacity at constant volume, (c) the final pressure and temperature, (d) the work done by the gas in the process and (e) the change in internal energy of the gas .
Answer» `PV=nRT` Given, `P=150 KPa =150 xx 10^3Pa`, `V= 150 cm^3 = 150 xx 10 ^(-6)m^3` `T=300K` (a `n=(PV)/(RT)=9.036 xx 10^(-3)` `=0.009` moles. (b) `(C_p)/(C_v) = gamma`, `C_p-C_v = R` So, `C_v = (R)/(gamma-1)=8.3/0.5 = 16.65//mol es`. (c) Given , `P_1 = 150 KPa = 150 xx 10^3 Pa`, `P_2 = ? V_1 = 150 cm^3` `=150 xx 10^(-6)m^3` `gamma=1.5` `V_2 = 50 cm^3 = 50 xx 10^(-6) m^3`, `T_1 = 300K`, `T_2=?` Since the process is adiabatic hence- `P_1V_1^gamma = P_2V_2^gamma` implies `150xx10^3 xx (150 xx 10^(-6))^gamma` `=P_2 xx ( 50 xx 10^(-6))^gamma` implies `P_2 = 150 xx 10^3 xx (150xx 10^6)^(1.5)/(50 xx 10 ^(-6))^1.5` `=150000xx(3)^1.5` `779.422 xx 10^8 Pa` `=780 KPa` Again, `P_1^(1-gamma) T_1^(gamma) = P_1^(1-gamma)T_2^(gamma)` implies `(150 xx 10^3)^(1-1.5) xx (330)^1.5` `=(780 xx 10 ^3) ^(1-1.5) xx T_2^1.5` implies `T_2^1.5 = (150 xx 10 ^3)^(1-1.5) xx (300)^1.5 xx 300 ^1.5` `=11849.050` implies `T_2= (11849.050)^(1//1.5)` `=519.74 = 520` (d) `dQ=W + dU` or ` W=-dU`[`dQ=0`, in adiabatic] `= -nCvdT` `=-0.009 xx 16.6 xx (520 -300)` = `-0.009 xx 16.6 xx 220` `=-32.8J = - 33J`. (e) `dU=nCvdT` `=0.009 xx 16.6 xx 220 = 33J`.

Discussion

7.	A gas is enclosed in a cylindrical can fitted with a piston. The walls of the can and the piston are adiabatic. The initial pressure , volume and temperture of the gas are `100 kpa, 400cm^(3)` and `300k` respectively. The ratio of the specific heat capacities of the gas is `(C_p / C_v = 1.5)` Find the pressure and the temperature of the gas if it is (a) suddenly compressed (b) slowly compressed to 100 cm^(3).
Answer» `P_1=100 Kpa` , `V_1 = 400 cm^3`, `=400xx10^(-6)m^3`, `T_1=300K`, `gamma=(Cp)/(Cv) = 1.5` (a) Suddenly compressed to `V_2 = 100 cm^3` `P_1V_1^gamma = P_2V_2^gamma` implies 10^5 xx (400)^1.5 = P_2(100)^1.5` `P_2 = 10^(5)(4)^1.5 = 800 KPa` `T_1V^(gamma-1)=T_2V_2^(gamma-1)` implies `300xx(400)^(1.5-1) = T_2(100)^(1.5-1)` implies `300 xx (400)^0.5 = T_2 (100)^0.5` `T_2 =600K`. (b) Even if the container is slowly compressed the walls are adiabatic so heat transferred is zero. Thus the values remain, `P_2=800 KPa` `T_2 =600 K`

Discussion

8.	A gas kept in a container of finite conductivity is suddenly compressed . The processA. must be very nearly adiabaticB. must be very nearly isothermalC. may be very nearly adiabaticD. may be very nearly isothermal
Answer» Correct Answer - C::D

Discussion

9.	`0.32 g` of exygen is kept in a rigid container and is heated. Find the amount of heat needed to raise the temperature from `25^0 C to 35^0 C`. The molar heat capacity of oxygen at constant volume is `20 J K^(-1) mol^(-1)`.
Answer» The molecular weight of oxygen `= 32 g mol^(-1)`. The amount of the gas in moles is `n=0.32 g / 32 g mol^(-1) = 0.01 mol`. The amount of heat needed is `Q = n C v Delta T (0.01 mol) (20 J K ^(-1) mol ^(-1) (10 K ) = 2.0 J`.

Discussion

10.	A sample of ideal gas `(gamma = 1.4)` is heated at constant pressure. If `140 J` of heat is supplied to gas, find `DeltaU` and W`.
Answer» Suppose the sample contains n moles. Also, suppose the volume changes from `V^(1) to V^(2)` and the temperature changes from `T^(1) to T^(2)`. The heat supplied is `DeltaQ = nCp(T^(2) -T^(1))`. (a) The change in internal energy is `DeltaU = nCv (T^(2) - T^(1)) = Cv / Cp nCP (T^(2) - T^(1))` `= Cv / Cp DeltaQ = 140 J / 1.4 = 100 J`. (b) The work done by the gas is `DeltaW = DeltaQ - DeltaU 140 J - 100 J = 40 J`.

Discussion

11.	A quantity of air is kept in a container having walls which are slightly conducting. The initial temperature and volume are `27^0C` (equal to the temperature of the surrounding) and `800cm^(3)` respectively. Find the rise in the temperature if the gas is compressed to `200 cm^(3) (a) in a short time (b) in a long time . Take gamma= 1.4.
Answer» when the gas is compressed in a short time, the process is adiabatic. Thus, `T_(2)V_(2)^(gamma-1) = T_(1)V_(1)^(gamma-1)` or, `T_(2) = T_(1)((V_1)/(V_2))^(gamma-1)` `=(300K)xx[(800)/(200)]^(0.4) = 522K` Rise in temperature = `T_(2) -T_(1) = 222 K` (b) When the gas is compressed in a long time. the process is isothermal thus, the temperature remains equal to the temperature of the sourrounding that is `27^(@)C`. THe rise in temperature = 0.

Discussion

12.	Two samples A and B of the same gas have equal volumes and pressures . The gas is sample A is expanded isothermally to double its volume and the the gas in B is expanded adibatically to double its volume . If the work done by the gas is the same for the two cases, show that gamma satisfies the equation`(1-2 ^(1-gamma)= (gamma - 1 ) 1n2`.
Answer» `P_1` = Initial pressure, `V_1` = Initial volume , `P_2` =Final pressure, `V_2`= Final volume Given ,`V_2 = 2V_1` Isothermal work done `=nRT_1l n (V_2)/(V_1)` Adiabatic work done `=(P_1V_1-P_2V_2)/(gamma-1)` Given that work done at both clases are same. Hence, `nRT_1l n = V_2/V_1` `=(P_1V_1-P_2V_2)/(gamma-1)` ...(1) At adiabatic process, `P_2 = P_1(V_2/V_1)^gamma=P_1(1/2)^gamma` From the equation (1) `nRT_1l n2=(P_1V_1(1-1/2^(gamma)2))/(gamma-1)` and `nRT_1 = P_1V_1` So `l n2 = (1-1/2^(gamma)2)/(gamma-1)` or `(gamma-1)l n2=1-2^(1-gamma)`

Discussion

13.	Calculate the internal energy of `1 g` of oxygen STP.
Answer» Oxygen is a diatomic gas. The average energy per molecule is, therefore, `5/2 kT`and the average energy per mole is 5/2RT. As the molecular weight of oxygen has `n= 1g / 32 g mol^(-1) = 1/ 32mol. The temperature of oxygen is 273 K. Thus, the internal energy is U `=n (5/2 RT)` `=( 1 /32 mol) (5/2) (8.31 J K^(-1) mol^(-1) (273 K ) 177 J`.

Discussion

14.	Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes. If `(C_A and C_B) be the molar heat capacities for the two processes,A. `C_A = C_B`B. `C_A lt C_B`C. `C_A gtC_B`D. `C_A and C_B` cannot be defined.
Answer» Correct Answer - C

Discussion

15.	Let `(Delta W_a) and (Delta W_b)` be the work done by the system A and B respectively in the previous question.A. `DeltaW_a gt DeltaW_b`B. `DeltaW_a = DeltaW_b`C. `DeltaW_a lt DeltaW_b`D. The relation between `DeltaW_a and DeltaW_b` cannot be deduced.
Answer» Correct Answer - C

Discussion

16.	A tank of volume `0.2 m^3` contains helium gas at a temperature of `300 K` and pressure `1.0 xx 10^5 N m^(-2)`. Find the amount of heat required to raise the temperature to `400 K`. The molar heat capacity of helium at constant volume is `3.0 cal K^(-1 ) mol^(-1)`. Neglact any expansion in the volume of tank.
Answer» The amount of the gas in moles is `n=p^V/RT` `=(1.0 xx 10^5 N m ^(-2)) (0.2 m ^3 )/(8.31 J K^(-1) mol^(-1)) (300 K )= 8.0mol`. The amount of heat required is `DeltaQ = nC_v Delta T` `=(8.0 mol) (3.0 cal mol^(-1) K^(-1)) (100k ) 2400 cal`.

Discussion

17.	The value of `(C_p - C_v)` is `1.00 R` for a gas sample in state A and is 1.08 R in state B. Let (`p_A, p_B)`denote the pressures and `(T_A and T_B)` denote the temperatures of the states A and B respectively . Most likelyA. `p_A lt p_B` and `T_A gt T_B`B. `p_A gt p_B and T_A lt T_B`C. `p_A = p_B and T_A lt T_B`D. `p_A gt p_B and T_A = T_B`
Answer» Correct Answer - A

Discussion

18.	The ratio of the molar heat capacities of an ideal gas is `(C_p / C_v 7/6)`. Calculate the change in internal energy of 1.0mole of the gas when its temperature is raised by `50 k (a)` keeping the pressure constant , (b) keeping the volume constant and (c ) adiabatically.
Answer» `(C_p/C_V)=7/6`, `n=1 mol.`, `DeltaT=50 K` (a) Keeping the pressure constant, `dQ = dU + dW`, `DeltaT = 50 K, gamma =7/6`, `n=1 mol.` `dQ=dU=dW` `implies nC_pdT=dU+RdT` `dU=nC_pdT-RdT` `=1xx(Rgamma)/(gamma-1)xxdT-RdT` `=7Rdt - RdT` `=7Rdt-RdT=6RdT` `=6xx8.3xx50 = 2490J.` (b) Keeping volume constant, `dU=nC_vdT` `=1 xx R/(gamma-1) xxdT` `=1((8.3)/(7//6)-1)xx50 ` `=8.3 xx 50 xx 6 = 2590 J.` (c) Adiabatically, `dQ=0`, `dU=-dW` `=[(nxxR)/(gamma-1)(T_1-T_2)]` `=(1xx83)/(7//6-1)= (T_2-T_1)` `=8.3 xx 6 xx 50 = 2490 J`.

Discussion

19.	Calculate the ratio `(C_p / C_v)` of oxygen from the following data . Speed of sound in oxygen`(= 32g mol^(-1))` and the gas constant `(R = 8.3 J K ^(-1) mol^(-1))`.
Answer» The speed of sound in a gas is given by `(v = (sqrt gamma P / rho )) = (sqrt gammaRT / M) ` or, `(gamma = M_v^(2)/ (RT) ` ` (32 xx 10^(-3) kg mol^(-1) (315 m s ^(-1) ) ^(2) / 8.3 J K ^(-1) mol^(-1) (273 K) = 1.4`

Discussion

20.	The molar heat capacity of a gas at constant volume is found to be `5 cal mol^(-1) K^(-1)`. Find the ratio`gamma = C_p/C_v` for the gas. The gas constant `R=2 cal mol^(-1) K^(-1).
Answer» We have `C_v = 5 cal mol^(-1) K^(-1). Thus. `C_p = C_v + R = 5 cal mol^(-1) K^(-1) + cal mol^(-1) = 7 cal mol^(-1) K^(-1)` or, C_p / C_v = 7/5 = 1.4`

Discussion

21.	The speed of sound in hydrogen at `0^@c` is `1280 m s^(-1)`. The density of hydrogen at STP is `0.089 kg m^(-3)`. Calculate the molar heat capacities `(C_p and C_v)` of hydrogen.
Answer» `V=1280 m//s`, `T=0^@C=273^@K` Density of `H_2 = 0.089 kg//m^3`, `R=8.3 J//mol-K` At STP, `P=10^5 Pa`, we know `V_(sound)= sqrt((gammap)/(rho))` `1280= sqrt((gamma xx 10 ^5)/(0.089))` or ` gamma= (1280 xx 1280 xx 0.089)/10^5` `=1.48` `C_p/C_v = gamma` or `C_p - C_v =R` `C_v = (R)/(gamma-1) = (8.31)/(1.48-1)` `=18.0 J//mol-k` `C_p = gammaC_v=1.48 xx 18.0` `=26.3//mol-K`.

Discussion

Explore topic-wise InterviewSolutions in .

The volume of an ideal gas `(gamma 1.5 )` is changed adiabatically from 4.00 liters to 3.00 liters . Find the ratio of (a) the final pressure to the initial pressure and (b) the final temperature to the initial temperature.A.B.C.D.

An ideal gas at pressure` 2.5 xx 10^(5)` `pa` and temperture `300k` occupies `100 cc`. It is adiabatically compressed to half its original volume. Calculate (a) the final pressure, (b) the final temperature and ( c) the work done by the gas in the peocess. Take `(gamma = 1.5)`.

Air `(gamma = 1.4 )` is pumped at 20atm pressure in a motor tyre at `20^@C`. If the tyre suddenly bursts, what would be the temperature of the air coming out of the tyre. Neglect any mixing with the atmoshpheric air.

A gas kept in a container of finite conductivity is suddenly compressed . The processA. must be very nearly adiabaticB. must be very nearly isothermalC. may be very nearly adiabaticD. may be very nearly isothermal

`0.32 g` of exygen is kept in a rigid container and is heated. Find the amount of heat needed to raise the temperature from `25^0 C to 35^0 C`. The molar heat capacity of oxygen at constant volume is `20 J K^(-1) mol^(-1)`.

A sample of ideal gas `(gamma = 1.4)` is heated at constant pressure. If `140 J` of heat is supplied to gas, find `DeltaU` and W`.

Calculate the internal energy of `1 g` of oxygen STP.

Let `(Delta W_a) and (Delta W_b)` be the work done by the system A and B respectively in the previous question.A. `DeltaW_a gt DeltaW_b`B. `DeltaW_a = DeltaW_b`C. `DeltaW_a lt DeltaW_b`D. The relation between `DeltaW_a and DeltaW_b` cannot be deduced.

The ratio of the molar heat capacities of an ideal gas is `(C_p / C_v 7/6)`. Calculate the change in internal energy of 1.0mole of the gas when its temperature is raised by `50 k (a)` keeping the pressure constant , (b) keeping the volume constant and (c ) adiabatically.

Calculate the ratio `(C_p / C_v)` of oxygen from the following data . Speed of sound in oxygen`(= 32g mol^(-1))` and the gas constant `(R = 8.3 J K ^(-1) mol^(-1))`.

The molar heat capacity of a gas at constant volume is found to be `5 cal mol^(-1) K^(-1)`. Find the ratio`gamma = C_p/C_v` for the gas. The gas constant `R=2 cal mol^(-1) K^(-1).

The speed of sound in hydrogen at `0^@c` is `1280 m s^(-1)`. The density of hydrogen at STP is `0.089 kg m^(-3)`. Calculate the molar heat capacities `(C_p and C_v)` of hydrogen.