Thermal Properties of Matter Test

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Thermal Properties of Matter Test - 25

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Question 1
1 / -0

The pressure of a gas is increased four times and its absolute temperature two times. The ratio of its final volume to its initial volume is

A
1:2

B
2:1

C
1:1

D
3:1

Solution

From ideal gas equation, PV = nRT
So, this gives $$V = \frac{nRT}{P}$$
so, V is proportional to absolute temperature and inversely proportional to the pressure of the gas
Now temperature is made four times and pressure is made 2 times
So, clearly the volume becomes half of it's original value.
So, A is the correct answer.
Question 2
1 / -0

A gas at 627$$^{0}$$ C is cooled so that its pressure becomes $$\dfrac{1}{3}$$ of its initial value at constant volume. Its final temperature is

A
900 K

B
600 K

C
300 K

D
100K

Solution

PV = nRT
since n and R and V are constant

$$P_{1}T_{2} = P_{2}T_{1}$$
here $$V_{1} = V_{2}$$
and $$T_{1} = 900$$
and $$P_{2} = \frac{P_{1}}{3}$$
and $$T_{2} = ?$$ (to be determined)
Putting all the values, we get $$T_{2}$$ = 300K
Question 3
1 / -0

At 27$$^{0}$$ C certain gas occupied a volume of 4 litres. If the volume of this is to be increased to 12 litres at the same pressure, the gas is to be heated to a temperature

A
300 K

B
900 K

C
600 K

D
1200K

Solution

PV = nRT
since n and R are constant, we get
$$\frac{PV}{T}$$ = constant
so, $$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$$
Here,
$$V_{1} = 4L$$
$$T_{1} = 300K$$
$$P_{2} = P_{1}$$
$$V_{2} = 12L$$
$$T_{2} = ? $$
putting all the values, we get :
$$T_{2}$$ = 900K
Question 4
1 / -0

A P-V diagram is obtained by changing the temperature of the gas as shown. During this process the gas is :

A
heated continuously

B
cooled continuously

C
heated in the beginning but cooled towards the end

D
cooled in the beginning but heated towards the end

Solution

From A to B
First initially pressure decrease and volume increases
equation of curve is $$P = -kV + c$$
Therefore, from ideal gas equation
$$T=PV/nR$$
$$T=(-kV+c)V/nR=(-kV^{2}+cV)/nR$$
Therefore its a parabola therefore in this process from A to B volume is increased therefore, its a parabola vertically downwards if we take V on x-axis and T on y-axis
Therefore as volume increases temperature rises becomes maximum and then again starts decreasing
Hence option(C)
Question 5
1 / -0

For an isochoric process the temperature at which the pressure of a gas will be double that of its pressure at 270$$^{0}$$ C is

A
543$$^{0}$$ C

B
813$$^{0}$$ C

C
3$$^{0}$$ C

D
270$$^{0}$$ C

Solution

PV = nRT
Since n and R are constant, we get
$$\frac{PV}{T}$$ = constant
so, $$\frac{P_{1}V_{1}}{T_{1}} = \frac{P_{2}V_{2}}{T_{2}}$$
Here,
$$T_{1} = 546K$$
$$P_{2} = 2P_{1}$$
$$V_{2} = V_{1}$$
$$T_{2} = ?$$
Putting all the values, we get : $$T_{2}$$ = 537$$^{\circ}$$ = 813K
Question 6
1 / -0

A car tyre has air at 1.5 atm at 300 K.If P increases to 1.75 atm with volume same, the temperature will be ____

A
350$$^{0}$$C

B
350K

C
300$$^{0}$$C

D
300K

Solution

HINT - Use ideal gas equation PV = nRT for the the cases.

$$\textbf{Step 1: Since n and R are constant, we get}$$
Use ideal gas equation PV = nRT
here n is number of moles.
P is pressure, V is volume, R is gas constant and T is temperature.
$$\dfrac{PV}{T}$$ = constant
so, $$\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}$$
Here,
$$P_{1} = 1.5atm$$
$$V_{1} = V_{2}$$
$$T_{1} = 300K$$
$$P_{2} = 1.75atm$$
$$T_{2} = ?$$
$$\textbf{Step 2: Putting all the values, we get : }$$
$$ T_2 = \left( \dfrac{P_2 V_2}{P_1 V_1 } \right) T_1$$

$$T_2=\dfrac{1.75}{1.5}\times 300$$

$$=350\ K$$

$${\textbf{Correct option: B}}$$
Question 7
1 / -0

The volume of a gas is 5 litres at N.T.P. what will be its volume at 273$$^{0}$$ C and at a pressure of four atmospheres

A
5 litres

B
2 litres

C
4 litres

D
2.5 litres

Solution

from ideal gas equation,
$$\frac{P_{1}V_{1}}{T_{1}}= \frac{P_{2}V_{2}}{T_{2}}$$
Here,
$$P_{1} = 1 atm$$
$$V_{1} = 5 L$$
$$T_{1} = 273 K$$
$$P_{2} = 4 atm$$
$$V_{2} = ?$$
$$T_{2} = 546 K$$
putting all the values, we get $$V_{2}$$ = 2.5 L
So, D is the correct answer.
Question 8
1 / -0

A gas at 27$$^{0}$$C and pressure of 30atm is allowed to expand to atmospheric pressure and volume 15 times larger. The final temperature of gas is

A
123$$^{0}$$C

B
-123$$^{0}$$C

C
400K

D
None

Solution

from ideal gas equation, PV = nRT
So, $$T = \frac{PV}{nR}$$
Since, n is constant, and R is also constant, we can say that T is directly proportional to the product of P and V
$$T \alpha PV$$
volume is made 15 times and pressure is reduced to $$\frac{1}{30}$$ th of it's original value (initial pressure = 30 atm and final pressure = 1 atm)
So, temperature becomes 15 x $$\frac{1}{30}$$ times = 0.5 times it's initial value
Initial temperature = 27$$^{\circ}$$C = 300K
So, final temperature = 0.5 x 300 K = 150 K = -123 $$^{\circ}$$C
So, B is the correct answer.
Question 9
1 / -0

At 20$$^{o}C$$ temperature and 1atmosphere pressure if a gas has a volume of 293 ml .Its volume at NTP is

A
$$546$$ cc

B
$$273$$ cc

C
$$293$$ cc

D
$$124$$ cc

Solution

from ideal gas equation,

$$\dfrac{P_{1}V_{1}}{T_{1}}= \dfrac{P_{2}V_{2}}{T_{2}}$$

Here,
$$P_{1} = 1 $$ atm

$$V_{1} = 293 $$ ml

$$T_{1} = 293$$ K

$$P_{2} = 1 $$ atm

$$V_{2} = ?$$

$$T_{2} = 273K$$

putting all the values, we get $$V_{2}$$ = 273 ml
Hence, option $$B$$ is the correct answer.
Question 10
1 / -0

A sample of an ideal gas occupies a volume V at pressure P and absolute temperature T. The mass of each molecule is m. The equation for density is

A
$$mKT$$

B
$$\dfrac{P}{KT}$$

C
$$\dfrac{P}{KTV}$$

D
$$\dfrac{Pm}{KT}$$

Solution

let us derive the expression for the density of an ideal gas
density = $$\dfrac{Mass}{Volume} = \dfrac{w}{V}$$
Now from ideal gas equation, $$PV = nRT$$
And n = $$\dfrac{w}{M}$$ where w is the mass of the gas and M is the molar mass of the gas
Putting this in ideal gas equation, we get
$$PV = \dfrac{w}{M}RT$$
Now substitute the value of V in the density expression to get
$$d = \dfrac{PM}{RT}$$
So, D is the correct answer.

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Thermal Properties of Matter Test - 25

Result

Thermal Properties of Matter Test - 25

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Question 1

1:2

2:1

1:1

3:1

Solution

Question 2

900 K

600 K

300 K

100K

Solution

Question 3

300 K

900 K

600 K

1200K

Solution

Question 4

heated continuously

cooled continuously

heated in the beginning but cooled towards the end

cooled in the beginning but heated towards the end

Solution

Question 5

543$$^{0}$$ C

813$$^{0}$$ C

3$$^{0}$$ C

270$$^{0}$$ C

Solution

Question 6

350$$^{0}$$C

350K

300$$^{0}$$C

300K

Solution

Question 7

5 litres

2 litres

4 litres

2.5 litres

Solution

Question 8

123$$^{0}$$C

-123$$^{0}$$C

400K

None

Solution

Question 9

$$546$$ cc

$$273$$ cc

$$293$$ cc

$$124$$ cc

Solution

Question 10

$$mKT$$

$$\dfrac{P}{KT}$$

$$\dfrac{P}{KTV}$$

$$\dfrac{Pm}{KT}$$

Solution

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