Advertisements
Advertisements
Question
A sample of an ideal gas (γ = 1.5) is compressed adiabatically from a volume of 150 cm3 to 50 cm3. The initial pressure and the initial temperature are 150 kPa and 300 K. 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.
Solution
The ideal gas equation is
PV = nRT
Given, P1 = 150 kPa = 150 × 103 Pa
V1 = 150 cm3 = 150 × 10−6 m3
T1= 300 K
(a)
`"n" =("P""V")/("R""T") = 9.036 xx 10^-3`
n = 0.009
(b)
`"C"_"p"/"C"_"v" = gamma , "C"_"p" -"C"_"v" ="R"`
So, `"C"_"v" = "R"/(gamma-1) = 2"R" = 8.3/0.5 = 16.6 "J" "mol" -"K"`
(c) Given,
P1 = 150 kPa = 150 × 103 Pa
P2 = ? V1 = 150 cm3
= 150 × 10−6 m3
γ = 1.5
V2 = 50 cm3 = 50 × 10−6 m3,
T1 = 300 K
T2 = ?
Since the process is adiabatic, using the equation of an adiabatic process,we get
P1V1γ = P2V2γ
⇒ 150 × 103 × (150 × 10−6)γ = P2 × (50 × 10−6)γ
`=> "P"_2 = 150 xx 10^3 xx ((150 xx 10^-6)^1.5)/(50 xx 10 ^-6)^1.5`
P2 = 150000 × (3)1.5
P2 = 779.422 × 103 Pa
P2 = 780 kPa
Again,
P11−γ T1γ = P11−γ T2γ
⇒ (150 × 103)1−1.5 × (330)1.5 = (780 × 103)1−1.5 × T21.5
⇒ T21.5 = (150 × 103)1−1.5 × (300)1.5 × 3001.5
T21.5 = 11849.050
⇒ T2 = (11849.050)1/1.5
T2 = 519.74 = 520 K
d) dQ = dW + dU
Or dW = −dU [ Since dQ = 0 in an adiabatic process]
dW = −nCvdT
dW = −0.009 × 16.6 × (520 − 300)
dW = −0.009 × 16.6 × 220
dW = −32.87 J ≈ −33 J
(e)
dU = nCvdT
dU = 0.009 × 16.6 × 220 ≈ 33 J
APPEARS IN
RELATED QUESTIONS
Consider a mixture of oxygen and hydrogen kept at room temperature. As compared to a hydrogen molecule an oxygen molecule hits the wall
Consider a sample of oxygen at 300 K. Find the average time taken by a molecule to travel a distance equal to the diameter of the earth.
Use R=8.314 JK-1 mol-1
Find the ratio of the mean speed of hydrogen molecules to the mean speed of nitrogen molecules in a sample containing a mixture of the two gases.
Use R = 8.314 JK-1 mol-1
Estimate the number of collisions per second suffered by a molecule in a sample of hydrogen at STP. The mean free path (average distance covered by a molecule between successive collisions) = 1.38 × 10−5 cm.
Use R = 8.31 JK−1 mol−1
Hydrogen gas is contained in a closed vessel at 1 atm (100 kPa) and 300 K. (a) Calculate the mean speed of the molecules. (b) Suppose the molecules strike the wall with this speed making an average angle of 45° with it. How many molecules strike each square metre of the wall per second?
Use R = 8.31 JK-1 mol-1
A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.
A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.
The ratio Cp / Cv for a gas is 1.29. What is the degree of freedom of the molecules of this gas?
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 CAand CB be the molar heat capacities for the two processes,
For a solid with a small expansion coefficient,
The value of Cp − Cv is 1.00 R for a gas sample in state A and 1.08 R in state B. Let pAand pB denote the pressures and TA and TB denote the temperatures of the states A and B, respectively. It is most likely that
Let Cv and Cp denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively. Which of the following is a universal constant?
70 calories of heat are required to raise the temperature of 2 mole of an ideal gas at constant pressure from 30° C to 35° C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is
The molar heat capacity for the process shown in the figure is
The molar heat capacity of oxygen gas at STP is nearly 2.5 R. As the temperature is increased, it gradually increases and approaches 3.5 R. The most appropriate reason for this behaviour is that at high temperatures
