An ideal gas of one mole is at temperature T and expands slowly according to the law P - V - constant- such that its final temperature is 15 T- If the work done by the gas is XRT- the value of X is -Answer upto two decimal points-

We know workdone by the gas is given by, ⇒ W = ∫ PdV It is given, P/V = k (constant). So, W = ∫ ^V_2 _V_1 kV dV = nbsp;k/2 [V_2^2 V_1 ^2] ....(i) As, ⇒P ...

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Standard XII

Physics

Problem Solving

An ideal gas ...

Question

An ideal gas of one mole is at temperature $T$ and expands slowly according to the law $\frac{P}{V} = c o n s t a n t$ , such that its final temperature is $15 T$ . If the work done by the gas is $x R T$ , the value of $x$ is
[Answer upto two decimal points]

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Solution

We know workdone by the gas is given by,
$\Rightarrow W = \int P d V$
It is given, $\frac{P}{V} = k (c o n s t a n t) .$
So, $W = \int_{V_{1}}^{V_{2}} k V d V = \frac{k}{2} [V_{2}^{2} - V_{1}^{2}] . . . . (i)$
As, $\Rightarrow \frac{P}{V} = k \Rightarrow \frac{n R T}{V^{2}} = k \Rightarrow \frac{T_{2}}{T_{1}} = {(\frac{V_{2}}{V_{1}})}^{2} \Rightarrow {(\frac{V_{2}}{V_{1}})}^{2} = 15 . . . . . (i i)$
From ( $i$ ) and ( $i i$ ),
$W = \frac{k}{2} [15 V_{1}^{2} - V_{1}^{2}] = \frac{14}{2} k V_{1}^{2} = 7 k V_{1}^{2} . . . . (i i i)$
We know, $P V = n R T \Rightarrow k V_{1}^{2} = 1 \times R T \Rightarrow k V_{1}^{2} = R T . . . . (i v)$
From ( $i i i$ ) and ( $i v$ ),
$\Rightarrow W = 7 R T, Hence x = 7.$

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An ideal gas of one mole is at temperature T and expands slowly according to the law PV=constant, such that its final temperature is 15T. If the work done by the gas is xRT, the value of x is [Answer upto two decimal points]

An ideal gas of one mole is at temperature $T$ and expands slowly according to the law $\frac{P}{V} = c o n s t a n t$ , such that its final temperature is $15 T$ . If the work done by the gas is $x R T$ , the value of $x$ is
[Answer upto two decimal points]