In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by . The amount of heat absorbed by gas is (R is gas co

Question

In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $Δ T$ . The amount of heat absorbed by gas is (R is gas constant):

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(1)

According to question $V T = K$

We also know that $P V = n R T$ $\Rightarrow T = (\frac{P V}{n R})$

$\Rightarrow V (\frac{P V}{n R}) = k \Rightarrow P V^{2} = K$

$C = \frac{R}{1 - 2} + \frac{3 R}{2} = \frac{R}{2}$

$∴$ $Δ Q = n C Δ T = \frac{R}{2} \times Δ T$ $[here, n = 1 mole]$

Solution

Q.
In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by $Δ T$ . The amount of heat absorbed by gas is (R is gas constant):

1 $\frac{1}{2} R Δ T$

2 $\frac{1}{2} K R Δ T$

3 $\frac{3}{2} R Δ T$

4 $\frac{2 K}{3} Δ T$

Ans.
(1)

According to question $V T = K$

We also know that $P V = n R T$ $\Rightarrow T = (\frac{P V}{n R})$

$\Rightarrow V (\frac{P V}{n R}) = k \Rightarrow P V^{2} = K$

$C = \frac{R}{1 - 2} + \frac{3 R}{2} = \frac{R}{2}$

$∴$ $Δ Q = n C Δ T = \frac{R}{2} \times Δ T$ $[here, n = 1 mole]$

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Solution

Q. In a process, temperature and volume of one mole of an ideal monoatomic gas are varied according to the relation VT = K, where K is a constant. In this process the temperature of the gas is increased by ΔT. The amount of heat absorbed by gas is (R is gas constant):

1 12RΔT

2 12KRΔT

3 32RΔT

4 2K3ΔT

Ans. (1) According to question VT=K We also know that PV=nRT⇒T=(PVnR) ⇒V(PVnR)=k⇒PV2=K C=R1-2+3R2=R2 ∴ ΔQ=nCΔT=R2×ΔT [here, n=1 mole]