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PHXI12:THERMODYNAMICS

371555 In a given process $d W = 0, d Q < 0$ then for a gas

1 Temperature increases

2 Volume decreases

3 Pressure decreases

4 Pressure increases

Explanation:

As $d W = 0$ so it is an isochoric process $\frac{P}{T} =$ constant
Given $d q = d U < 0 \Rightarrow T$ decreases and hence $P$ also decreases.

PHXI12:THERMODYNAMICS

371556 If $R =$ universal gas constant, the amount of heat needed to rise the temperature of $2 m o l$ of an ideal monoatomic gas from $273 K$ to $373 K$ when no work is done is

1

100 R

2

150 R

3

300 R

4

500 R

Explanation:

$Δ Q = Δ U = n C_{V} Δ T = 2 \times \frac{3 R}{2} \times 100 = 300 R$

PHXI12:THERMODYNAMICS

371557 In an isochoric process, the correct ratio is

1

Δ Q : W = 1 : 1

2

Δ Q : W = γ : γ - 1

3

Δ Q : W = γ - 1 : γ

4

Δ Q : W = γ : 1

Explanation:

$\frac{Δ Q}{W} = \frac{Δ Q}{Δ Q - Δ U} = \frac{C_{P}}{C_{P} - C_{V}} = \frac{γ}{(γ - 1)}$

PHXI12:THERMODYNAMICS

371558 Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes $V_{1}$ and $V_{2}$ , then:
supporting img

1

V_{1} > V_{2}

2

V_{1} < V_{2}

3

V_{1} = V_{2}

4 The information is insufficient

Explanation:

Use $P V = n R T$
$\frac{P}{T} = \frac{n R}{V}$
$∴ s l o p e \propto \frac{1}{V}$
So correct Option is (1)

PHXI12:THERMODYNAMICS

371559 $0.08 k g$ air is heated at constant volume through $5^{\circ} C$ . The specific heat of air at constant volume is $0.17 k c a l / k g^{\circ} C$ and $J = 4.18 j o u l e / c a l$ . The change in its internal energy is approximately.

1

298 J

2

284 J

3

318 J

4

142 J

Explanation:

$Δ U = m C_{V} Δ t$
$(∵ Δ U = \frac{t}{2} n R Δ T and C_{v} = \frac{t}{2} R)$
$= 0.08 \times 0.17 \times 10^{3} \times 5$
$= 0.068 \times 10^{3} c a l$
$= 68 c a l = 284.2 J$

JEE - 2024

PHXI12:THERMODYNAMICS

371555 In a given process $d W = 0, d Q < 0$ then for a gas

1 Temperature increases

2 Volume decreases

3 Pressure decreases

4 Pressure increases

Explanation:

As $d W = 0$ so it is an isochoric process $\frac{P}{T} =$ constant
Given $d q = d U < 0 \Rightarrow T$ decreases and hence $P$ also decreases.

PHXI12:THERMODYNAMICS

371556 If $R =$ universal gas constant, the amount of heat needed to rise the temperature of $2 m o l$ of an ideal monoatomic gas from $273 K$ to $373 K$ when no work is done is

1

100 R

2

150 R

3

300 R

4

500 R

Explanation:

$Δ Q = Δ U = n C_{V} Δ T = 2 \times \frac{3 R}{2} \times 100 = 300 R$

PHXI12:THERMODYNAMICS

371557 In an isochoric process, the correct ratio is

1

Δ Q : W = 1 : 1

2

Δ Q : W = γ : γ - 1

3

Δ Q : W = γ - 1 : γ

4

Δ Q : W = γ : 1

Explanation:

$\frac{Δ Q}{W} = \frac{Δ Q}{Δ Q - Δ U} = \frac{C_{P}}{C_{P} - C_{V}} = \frac{γ}{(γ - 1)}$

PHXI12:THERMODYNAMICS

371558 Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes $V_{1}$ and $V_{2}$ , then:
supporting img

1

V_{1} > V_{2}

2

V_{1} < V_{2}

3

V_{1} = V_{2}

4 The information is insufficient

Explanation:

Use $P V = n R T$
$\frac{P}{T} = \frac{n R}{V}$
$∴ s l o p e \propto \frac{1}{V}$
So correct Option is (1)

PHXI12:THERMODYNAMICS

371559 $0.08 k g$ air is heated at constant volume through $5^{\circ} C$ . The specific heat of air at constant volume is $0.17 k c a l / k g^{\circ} C$ and $J = 4.18 j o u l e / c a l$ . The change in its internal energy is approximately.

1

298 J

2

284 J

3

318 J

4

142 J

Explanation:

$Δ U = m C_{V} Δ t$
$(∵ Δ U = \frac{t}{2} n R Δ T and C_{v} = \frac{t}{2} R)$
$= 0.08 \times 0.17 \times 10^{3} \times 5$
$= 0.068 \times 10^{3} c a l$
$= 68 c a l = 284.2 J$

JEE - 2024

PHXI12:THERMODYNAMICS

371555 In a given process $d W = 0, d Q < 0$ then for a gas

1 Temperature increases

2 Volume decreases

3 Pressure decreases

4 Pressure increases

Explanation:

As $d W = 0$ so it is an isochoric process $\frac{P}{T} =$ constant
Given $d q = d U < 0 \Rightarrow T$ decreases and hence $P$ also decreases.

PHXI12:THERMODYNAMICS

371556 If $R =$ universal gas constant, the amount of heat needed to rise the temperature of $2 m o l$ of an ideal monoatomic gas from $273 K$ to $373 K$ when no work is done is

1

100 R

2

150 R

3

300 R

4

500 R

Explanation:

$Δ Q = Δ U = n C_{V} Δ T = 2 \times \frac{3 R}{2} \times 100 = 300 R$

PHXI12:THERMODYNAMICS

371557 In an isochoric process, the correct ratio is

1

Δ Q : W = 1 : 1

2

Δ Q : W = γ : γ - 1

3

Δ Q : W = γ - 1 : γ

4

Δ Q : W = γ : 1

Explanation:

$\frac{Δ Q}{W} = \frac{Δ Q}{Δ Q - Δ U} = \frac{C_{P}}{C_{P} - C_{V}} = \frac{γ}{(γ - 1)}$

PHXI12:THERMODYNAMICS

371558 Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes $V_{1}$ and $V_{2}$ , then:
supporting img

1

V_{1} > V_{2}

2

V_{1} < V_{2}

3

V_{1} = V_{2}

4 The information is insufficient

Explanation:

Use $P V = n R T$
$\frac{P}{T} = \frac{n R}{V}$
$∴ s l o p e \propto \frac{1}{V}$
So correct Option is (1)

PHXI12:THERMODYNAMICS

371559 $0.08 k g$ air is heated at constant volume through $5^{\circ} C$ . The specific heat of air at constant volume is $0.17 k c a l / k g^{\circ} C$ and $J = 4.18 j o u l e / c a l$ . The change in its internal energy is approximately.

1

298 J

2

284 J

3

318 J

4

142 J

Explanation:

$Δ U = m C_{V} Δ t$
$(∵ Δ U = \frac{t}{2} n R Δ T and C_{v} = \frac{t}{2} R)$
$= 0.08 \times 0.17 \times 10^{3} \times 5$
$= 0.068 \times 10^{3} c a l$
$= 68 c a l = 284.2 J$

JEE - 2024

PHXI12:THERMODYNAMICS

371555 In a given process $d W = 0, d Q < 0$ then for a gas

1 Temperature increases

2 Volume decreases

3 Pressure decreases

4 Pressure increases

Explanation:

As $d W = 0$ so it is an isochoric process $\frac{P}{T} =$ constant
Given $d q = d U < 0 \Rightarrow T$ decreases and hence $P$ also decreases.

PHXI12:THERMODYNAMICS

371556 If $R =$ universal gas constant, the amount of heat needed to rise the temperature of $2 m o l$ of an ideal monoatomic gas from $273 K$ to $373 K$ when no work is done is

1

100 R

2

150 R

3

300 R

4

500 R

Explanation:

$Δ Q = Δ U = n C_{V} Δ T = 2 \times \frac{3 R}{2} \times 100 = 300 R$

PHXI12:THERMODYNAMICS

371557 In an isochoric process, the correct ratio is

1

Δ Q : W = 1 : 1

2

Δ Q : W = γ : γ - 1

3

Δ Q : W = γ - 1 : γ

4

Δ Q : W = γ : 1

Explanation:

$\frac{Δ Q}{W} = \frac{Δ Q}{Δ Q - Δ U} = \frac{C_{P}}{C_{P} - C_{V}} = \frac{γ}{(γ - 1)}$

PHXI12:THERMODYNAMICS

371558 Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes $V_{1}$ and $V_{2}$ , then:
supporting img

1

V_{1} > V_{2}

2

V_{1} < V_{2}

3

V_{1} = V_{2}

4 The information is insufficient

Explanation:

Use $P V = n R T$
$\frac{P}{T} = \frac{n R}{V}$
$∴ s l o p e \propto \frac{1}{V}$
So correct Option is (1)

PHXI12:THERMODYNAMICS

371559 $0.08 k g$ air is heated at constant volume through $5^{\circ} C$ . The specific heat of air at constant volume is $0.17 k c a l / k g^{\circ} C$ and $J = 4.18 j o u l e / c a l$ . The change in its internal energy is approximately.

1

298 J

2

284 J

3

318 J

4

142 J

Explanation:

$Δ U = m C_{V} Δ t$
$(∵ Δ U = \frac{t}{2} n R Δ T and C_{v} = \frac{t}{2} R)$
$= 0.08 \times 0.17 \times 10^{3} \times 5$
$= 0.068 \times 10^{3} c a l$
$= 68 c a l = 284.2 J$

JEE - 2024

PHXI12:THERMODYNAMICS

371555 In a given process $d W = 0, d Q < 0$ then for a gas

1 Temperature increases

2 Volume decreases

3 Pressure decreases

4 Pressure increases

Explanation:

As $d W = 0$ so it is an isochoric process $\frac{P}{T} =$ constant
Given $d q = d U < 0 \Rightarrow T$ decreases and hence $P$ also decreases.

PHXI12:THERMODYNAMICS

371556 If $R =$ universal gas constant, the amount of heat needed to rise the temperature of $2 m o l$ of an ideal monoatomic gas from $273 K$ to $373 K$ when no work is done is

1

100 R

2

150 R

3

300 R

4

500 R

Explanation:

$Δ Q = Δ U = n C_{V} Δ T = 2 \times \frac{3 R}{2} \times 100 = 300 R$

PHXI12:THERMODYNAMICS

371557 In an isochoric process, the correct ratio is

1

Δ Q : W = 1 : 1

2

Δ Q : W = γ : γ - 1

3

Δ Q : W = γ - 1 : γ

4

Δ Q : W = γ : 1

Explanation:

$\frac{Δ Q}{W} = \frac{Δ Q}{Δ Q - Δ U} = \frac{C_{P}}{C_{P} - C_{V}} = \frac{γ}{(γ - 1)}$

PHXI12:THERMODYNAMICS

371558 Figure shows the pressure versus temperature curves for a given mass of a gas corresponding to two different volumes $V_{1}$ and $V_{2}$ , then:
supporting img

1

V_{1} > V_{2}

2

V_{1} < V_{2}

3

V_{1} = V_{2}

4 The information is insufficient

Explanation:

Use $P V = n R T$
$\frac{P}{T} = \frac{n R}{V}$
$∴ s l o p e \propto \frac{1}{V}$
So correct Option is (1)

PHXI12:THERMODYNAMICS

371559 $0.08 k g$ air is heated at constant volume through $5^{\circ} C$ . The specific heat of air at constant volume is $0.17 k c a l / k g^{\circ} C$ and $J = 4.18 j o u l e / c a l$ . The change in its internal energy is approximately.

1

298 J

2

284 J

3

318 J

4

142 J

Explanation:

$Δ U = m C_{V} Δ t$
$(∵ Δ U = \frac{t}{2} n R Δ T and C_{v} = \frac{t}{2} R)$
$= 0.08 \times 0.17 \times 10^{3} \times 5$
$= 0.068 \times 10^{3} c a l$
$= 68 c a l = 284.2 J$

JEE - 2024

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